Chapter 5.23 - Solved examples on Centripetal Force

In the previous section, we saw centripetal force. In this section, we will see some solved examples.

Solved example 5.40
A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string ? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N ?
Solution:
Part (a):
1. Speed of the stone = 40 rev./min = $\mathbf\small{\frac{40}{60}}$ rev./sec = $\mathbf\small{\frac{2}{3}}$ rev./sec
• That means, the stone completes 'two third of a revolution' in one second
• That means, the stone turns through $\mathbf\small{(\frac{2}{3}\times 2 \pi)=\frac{4 \pi}{3}}$ radians in one second
• Thus we get: Angular velocity of the stone $\mathbf\small{\omega=\frac{4 \pi}{3}}$ rad/sec
2. Centripetal force required = $\mathbf\small{|\vec{f_c}|=\frac{m|\vec{v}|^2}{R}=\frac{mR^2 \omega^2}{R}=mR\omega^2}$
($\mathbf\small{\because |\vec{v}|=R\omega}$)
Substituting the values, we get: $\mathbf\small{|\vec{f_c}|=0.25 \times 1.5 \times \left\lgroup\frac{4 \pi}{3}\right\rgroup ^2=6.57\, \text{N}}$
3. The required centripetal force for the stone is provided by the tension in the string. 
Thus we get: Tension in the string = 6.57 N
Part (b):
1. Let the maximum allowable speed be $\mathbf\small{\omega_{max}}$ 
• Then we can write: $\mathbf\small{|\vec{f_c}|=200 \, \text{N}=mR\omega_{max}^2}$
• Substituting the values, we get: $\mathbf\small{200 =0.25 \times 1.5 \times\omega_{max}^2}$
• Thus we get: $\mathbf\small{\omega_{max}}$ = 23.094 rad/sec
2. So the stone can cover 23.094 radians in 1 second
• Then number of revolutions in 1 second = $\mathbf\small{\frac{23.094}{2\pi}=3.68}$
• We can express the maximum allowable speed in terms of linear velocity also. We have:
$\mathbf\small{|\vec{v_{max}}|=R\omega_{max}=1.5\times23.094=34.64\,\text{ms}^{-1}}$

Solved example 5.41
A disc revolves with a speed of 33¹⁄₃ rev/min, and has a radius of 15 cm. Two coins are placed at 4 cm and 14 cm away from the centre of the record. If the co-efficient of friction between the coins and the record is 0.15, which of the coins will revolve with the record ?
Solution:
The arrangement is shown in fig.5.92(a) below:

Fig.5.92

1. Angular speed of the disc = 33¹⁄₃ rev./min = ¹⁰⁰⁄₃ rev./min = $\mathbf\small{\frac{100}{180}}$ rev./sec = $\mathbf\small{\frac{5}{9}}$ rev./sec
• That means, the disc turns through $\mathbf\small{(\frac{5}{9}\times 2 \pi)=\frac{10 \pi}{9}}$ radians in one second
• Thus we get: Angular speed of the disc $\mathbf\small{\omega=\frac{10 \pi}{9}}$ rad/sec
2. Though the angular speed of the two coins are the same, their linear speeds will be different (Details here)
• Linear speed of coin 1 = $\mathbf\small{|\vec{v_{c1}}|=R_1\omega=0.04\times\frac{10 \pi}{9}=\frac{0.4 \pi}{9}}$
• Linear speed of coin 2 = $\mathbf\small{|\vec{v_{c2}}|=R_2\omega=0.14\times\frac{10 \pi}{9}=\frac{1.4 \pi}{9}}$
3. Centripetal force required for the first coin =
$\mathbf\small{|\vec{f_{c1}}|=\frac{m|\vec{v_1}|^2}{R_1}=m \times \frac{1}{0.04} \times \left\lgroup\frac{0.4 \pi}{9}\right\rgroup ^2=\frac{4\pi^2m}{81}=0.487m\, \text{N}}$ 
• Centripetal force required for the second coin =
$\mathbf\small{|\vec{f_{c2}}|=\frac{m|\vec{v_2}|^2}{R_2}=m \times \frac{1}{0.14} \times \left\lgroup\frac{1.4 \pi}{9}\right\rgroup ^2=\frac{14\pi^2m}{81}=1.706m\, \text{N}}$
4. For both the coins, the centripetal force will be provided by the friction
Magnitude of the frictional force for each coin = $\mathbf\small{\mu_smg=0.15 \times m \times 10=1.5m \, \text{N}}$
• Comparing with the results in (3), we see that:
(i) The centripetal force required for the inner coin is less than the force available from friction
(ii) The centripetal force required for the outer coin is more than the force available from friction
■ So we can write: 
• The inner coin will revolve with the disc
• The outer coin will slip away

Solved example 5.42
A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical axis with 200 rev/min. See fig.5.92(b) above. The coefficient of friction between the wall and his clothing is 0.15. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed ?
Solution:
1. When the man stands inside a stationary hollow cylinder, his weight is supported by the floor of the cylinder
• When the cylinder rotates, he will be pressed against the inner surface of the cylinder. This pressing is due to the centripetal force
• Because of this pressing, frictional force will develop between his clothes and the wall of the cylinder
• Our first step is to find the magnitude of this frictional force
We have:
Static frictional force = $\mathbf\small{|\vec{f_s}|=\mu_s|\vec{F_N}|=0.15 \times |\vec{F_N}|}$
2. But $\mathbf\small{|\vec{F_N}|}$ =Centripetal force = $\mathbf\small{|\vec{f_c}|=mR\omega_{min}^2}$
• Where $\mathbf\small{\omega_{min}}$ is the minimum angular velocity required
• Substituting the known values, we get: $\mathbf\small{|\vec{f_c}|=m \times 3 \times \left\lgroup\omega_{min}\right\rgroup ^2}$
3. We can put this value of $\mathbf\small{|\vec{f_c}|}$ in the place of $\mathbf\small{|\vec{F_N}|}$ in (1). We get:
$\mathbf\small{|\vec{f_s}|=0.15 \times m \times 3 \times \left\lgroup\omega_{min}\right\rgroup ^2}$
4. The above result in (3) is the static friction which holds the man from falling down when the floor is removed
• So this friction must be equal to 'mg'. We can write:
$\mathbf\small{0.15 \times m \times 3 \times \left\lgroup\omega_{min}\right\rgroup ^2=mg}$
$\mathbf\small{\Longrightarrow \omega_{min}= \sqrt{\frac{g}{0.15 \times 3}}=4.71\,\text{rad/sec}}$

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In the next chapter, we will see work, energy and power

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Chapter 5.22 - The Centripetal Force

In the previous section, we saw rolling friction. In this section, we will see centripetal force.

• Consider fig.5.88(a) below. A small object of mass ‘m’ is tied to one end of an inextensible string.

Fig.5.88

• A person holds the other end of the string and whirls the object around in a horizontal plane.
    ♦ Obviously, the path followed by the object will be a circle. 
    ♦ The radius of that circle will be equal to the length of the string. 
• Let the person whirl it in such a way that, the motion is uniform. This is shown in fig.b 
• We have seen 'uniform circular motion' in a previous section. 
• We want to know the forces acting on the object which is under a 'uniform circular motion'. 
• We will write the analysis in steps:
1. When the object moves around the circle, the string will be in tension. 
• Remember that, a string can take only tension. It cannot take compression. We have seen the details in a previous section. 
2. So we can draw two opposite arrows on the string. This is shown in fig.c. 
• The two arrows tell us these:
    ♦ The string pulls at the hand
    ♦ The string pulls at the object
3. Let us choose the object as a sub-system. This is shown by the red rectangle in fig.c
• The FBD will be as shown in fig.d
• We see that the tension in the string is pulling the object towards the center of the circle
4. We know that FBDs show only those forces which are experienced by the sub-systems.
• They do not show forces which are applied by the sub-systems
• So we can write: The object experiences a pull $\mathbf\small{\vec{T}}$ towards the center
5. Let the object be at any position along the circle. We can draw the FBD at that position.
• Wherever (along the circle) be the position of the object, We will see the same $\mathbf\small{\vec{T}}$ in its FBD.
• That means, the object is subjected to a constant pull $\mathbf\small{\vec{T}}$ towards the center.
6. If we cut the string while the circular motion is on, the $\mathbf\small{\vec{T}}$ will disappear. (see fig.5.89.a below)

Fig.5.89

• This is because, when the string is cut, the medium to provide $\mathbf\small{\vec{T}}$ is no longer available
7. The object will then fly off in a straight line.
• This straight line will be tangential to the circle.
• The ‘point of tangency’ is the ‘position of the object at the exact instant when the cut was made’. This is shown in fig.5.89(b) above.
■ So we see that a ‘force towards the center’ is essential to maintain circular motion.
• We want the details of this force.
8. Earlier, we have seen centripetal acceleration $\mathbf\small{\vec{a_c}}$ (Details here)
• Its magnitude is given by $\mathbf\small{|\vec{a_c}|=\frac{v^2}{R}}$
    ♦ Where v is the speed
    ♦ R is the radius of the circle
• Its direction changes continuously, being always directed towards the center
9. By the second law, any object subjected to acceleration will experience a force $\mathbf\small{\vec{F}}$
• The magnitude of this force is given by $\mathbf\small{|\vec{F}|=\text{mass}\,\times\,|\vec{a}|}$
10. Now, our object is also subjected to acceleration $\mathbf\small{\vec{a_c}}$
• So a force must be acting on it. It is known as the centripetal force and is denoted as $\mathbf\small{\vec{f_c}}$
• Obviously, the magnitude of $\mathbf\small{\vec{f_c}}$ will be given by:
Eq.5.5:  $\mathbf\small{|\vec{f_c}|=\frac{mv^2}{R} }$
11. The direction of $\mathbf\small{\vec{f_c}}$ will be the same as the direction of $\mathbf\small{\vec{a_c}}$
• That means the direction of the $\mathbf\small{\vec{f_c}}$ is changing continuously. It is always directed towards the center
12. So in fig.5.88(d) above, $\mathbf\small{|\vec{T}|=|\vec{f_c}|}$
• No object can perform a circular motion without the help of $\mathbf\small{\vec{f_c}}$
• In fact, if an object has to move along even a small part of a curve, the centripetal force is to be provided
■ For planets moving around the sun, the centripetal force is provided by the gravitational force
■ What about a car moving along a curved path? We will see it in detail:

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The motion of a car along a circular road
Two cases come under this topic:
Case 1: The road is level
Case 2: The road  is banked
We will see each case in detail:
Case 1: The motion of a car on a level road
1. Fig.5.90(a) below shows a car moving along a circular road. 
• It is moving along the center-line of the road. The center-line is shown using a dashed curve.

Fig.5.90

• The radius of the center-line of the road is 'R'.
2. Now, centripetal force must be provided for the car so that it can perform the circular motion.
■ Where does this force come from?
• The forces acting on the car are shown in fig.b
• The vertical forces will cancel each other. 
• There will be a frictional force, which is horizontal. 
• This force acts at the interface between the tire and the surface of the road.
■ It is this frictional force that provides the necessary centripetal force.
3. We know that there are two types of friction: static and kinetic
• It is the static friction that provides the centripetal force. 
• The reason can be written in the following steps (i) to (iv) 
(i) The car will be always trying to move off in a linear direction
(ii) But friction keeps the car in the circular path
(iii) That means friction opposes the impending linear motion. 
(iv) So it is the static friction which is playing the role here.  
4. We know that 'external force opposing the static frictional force' must be always less than or equal to $\mathbf\small{\mu_s |\vec{F_N}|}$
Otherwise, the static frictional force will no longer be able to hold (Details here)
5. In our present case, the force which opposes the static frictional force is the centripetal force $\mathbf\small{\vec{f_c}}$
So we can write: $\mathbf\small{|\vec{f_c}|\leq \mu_s |\vec{F_N}|}$
$\mathbf\small{\Longrightarrow \frac{mv^2}{R}\leq \mu_s |\vec{F_N}|}$
$\mathbf\small{\Longrightarrow \frac{mv^2}{R}\leq \mu_s \times mg}$
$\mathbf\small{\Longrightarrow \frac{v^2}{R}\leq \mu_s \times g}$
$\mathbf\small{\Longrightarrow {v^2}\leq \mu_s \times Rg}$
$\mathbf\small{\Longrightarrow {v}\leq \sqrt{\mu_s  Rg}}$
6. In the final expression that we obtained in the step (5) above, μ_s, R and g are constants.
• 'v' is the only variable. This is obvious because a car can move with variable speeds. The radius of the road will remain the same
• So it is important to keep the speed less than or equal to $\mathbf\small{\sqrt{\mu_s  Rg}}$
• Otherwise, the friction capacity will be exceeded. The tire will not be able to provide the necessary centripetal force. And the car will skid off from the circular path causing an accident.
7. So drivers must be warned about the maximum speed they can attain on a curved road
• From the result in step (5), we see that the maximum allowable speed is $\mathbf\small{\sqrt{\mu_s  Rg}}$
• We can write:
Eq.5.6: $\mathbf\small{v_{max}=\sqrt{\mu_s  Rg}}$
• This maximum allowable speed is displayed on the roadsides in such a way that drivers from both directions see them much before they reach the curve. Some examples can be seen here.    
• From Eq.5.6, we see that $\mathbf\small{v_{max}}$ is independent of the mass of the car. That means the $\mathbf\small{v_{max}}$ obtained is applicable to heavy vehicles and light vehicles alike
• The factors on which $\mathbf\small{v_{max}}$ depends on are: μ_s, R and g 
• There are, however, many more factors to be considered before finalizing the value of $\mathbf\small{v_{max}}$. Those factors are detailed in Highway Engineering textbooks.

Case 2: The motion of a car on a banked road
1. In this case, the outer edge of the road is raised to a higher level than the inner edge. This is shown in fig.5.91(a) below:

Fig.5.91

2. We see that the force of friction $\mathbf\small{\vec{f_s}}$ is now inclined. 
• This is obvious because frictional force will always be parallel to the surface of contact. 
3. The other two forces are: 
• $\mathbf\small{\vec{W_{car}}}$ acting vertically downwards
• $\mathbf\small{\vec{F_N}}$ acting perpendicular to the surface of the road.
4. For our present analysis, we do not want any inclined forces. All forces that we bring into calculations should be either vertical or horizontal.
• $\mathbf\small{\vec{W_{car}}}$ is already vertical
• So we will resolve the inclined forces $\mathbf\small{\vec{F_N}}$ and $\mathbf\small{\vec{f_s}}$ into their rectangular components.
5. First, we take $\mathbf\small{\vec{F_N}}$. See fig.5.91(b)
(i) Draw horizontal and vertical dashed lines through the tail end of $\mathbf\small{\vec{F_N}}$
(ii) Identify the 'position of θ'. 
• By the properties of right triangles, we will see that θ is between $\mathbf\small{\vec{F_N}}$ and the vertical
(iii) Once θ is fixed, we can easily write the components (Details here):
    ♦ The component which is adjacent to the angle gets the cosine
    ♦ The other component gets the sine
■ Thus we get:
• The horizontal component as: $\mathbf\small{(|\vec{F_N}|\sin \theta)\hat{i}}$
• The vertical component as: $\mathbf\small{(|\vec{F_N}|\cos \theta)\hat{j}}$
• They are shown in fig.c
6. Next, we take $\mathbf\small{\vec{f_s}}$. See fig.5.91(d)
(i) Draw horizontal and vertical dashed lines through the tail end of $\mathbf\small{\vec{f_s}}$
(ii) Identify the 'position of θ'. 
• By the properties of right triangles, we will see that θ is between $\mathbf\small{\vec{f_s}}$ and the horizontal
(iii) Once θ is fixed, we can easily write the components:
    ♦ The component which is adjacent to the angle gets the cosine
    ♦ The other component gets the sine
■ Thus we get:
• The horizontal component as: $\mathbf\small{(|\vec{f_s}|\cos \theta)\hat{i}}$
• The vertical component as: $\mathbf\small{(|\vec{f_s}|\sin \theta)\hat{j}}$
• They are shown in fig.e
7. So both the inclined forces are now resolved into their respective components.
• In fig.f, those components are arranged in order. So now we can see their effects:  
(i) There is no motion in the vertical direction. So the vertical components can be related as:
$\mathbf\small{|\vec{F_N}|\cos \theta=|\vec{f_s}|\sin \theta+mg}$
(ii) The two horizontal components together provide the necessary centripetal force. So we can write:
$\mathbf\small{|\vec{F_N}|\sin \theta+|\vec{f_s}|\cos \theta=\frac{mv^2}{R}}$
8. So we see that, both $\mathbf\small{\vec{F_N}}$ and $\mathbf\small{\vec{f_s}}$ contributes towards providing the centripetal force
• Now, the maximum contribution that can be expected from $\mathbf\small{\vec{f_s}}$ is $\mathbf\small{\mu_s |\vec{F_N}|}$
• When this maximum possible contribution is obtained, the car will be able to move with the maximum allowable velocity $\mathbf\small{v_{max}}$
• So the result in 7(ii) can be written as: $\mathbf\small{|\vec{F_N}|\sin \theta+\mu_s|\vec{F_N}|\cos \theta=\frac{mv_{max}^2}{R}}$
$\mathbf\small{\Longrightarrow|\vec{F_N}|(\sin \theta+\mu_s\cos \theta)=\frac{mv_{max}^2}{R}}$
9. The previous result in 7(i) can also be rewritten in terms of μ_s
$\mathbf\small{|\vec{F_N}|\cos \theta=\mu_s|\vec{F_N}|\sin \theta+mg}$
$\mathbf\small{\Longrightarrow|\vec{F_N}|(\cos \theta-\mu_s\sin \theta)=mg}$
$\mathbf\small{\Longrightarrow|\vec{F_N}|=\frac{mg}{(\cos \theta-\mu_s\sin \theta)}}$
10. In (9) above, we get an expression for $\mathbf\small{|\vec{F_N}|}$. We can use this expression in the place of $\mathbf\small{|\vec{F_N}|}$ in (8). We get:
$\mathbf\small{\frac{mg}{(\cos \theta-\mu_s\sin \theta)}\times(\sin \theta+\mu_s\cos \theta)=\frac{mv_{max}^2}{R}}$
$\mathbf\small{\Longrightarrow\frac{g(\sin \theta+\mu_s\cos \theta)}{(\cos \theta-\mu_s\sin \theta)}=\frac{v_{max}^2}{R}}$
• Dividing numerator and denominator by cos θ, we get:
$\mathbf\small{\frac{g(\mu_s+\tan \theta)}{(1-\mu_s\tan \theta)}=\frac{v_{max}^2}{R}}$
$\mathbf\small{\Longrightarrow {v_{max}^2}=\frac{Rg(\mu_s+\tan \theta)}{(1-\mu_s\tan \theta)}}$
• Thus we get Eq.5.7: $\mathbf\small{ {v_{max}}=\left[\frac{Rg(\mu_s+\tan \theta)}{1-\mu_s\tan \theta}\right]^{\frac{1}{2}}}$
11. If the banked road is friction less, the required centripetal force will have to be provided by the 'banking effect' alone. There will not be any contribution from the friction
Let us write the steps briefly:
(i) Fig.5.91(f) will have only two vertical forces:
• $\mathbf\small{(|\vec{F_N}|\cos \theta)\hat{j}}$ (upwards)
• $\mathbf\small{\vec{W_{car}}}$ (downwards)
(ii) There is no motion in the vertical direction. So we can write:
$\mathbf\small{|\vec{F_N}|\cos \theta=mg}$
(iii) Also, fig.5.91(f) will have only one horizontal force:
• $\mathbf\small{|\vec{F_N}|\sin \theta}$
(iv) This horizontal component provide the necessary centripetal force. So we can write:
$\mathbf\small{|\vec{F_N}|\sin \theta=\frac{mv_0^2}{R}}$
• Where v₀ is the maximum velocity of the car when no contribution from friction is received
(v) From (ii) we have: $\mathbf\small{|\vec{F_N}|=\frac{mg}{\cos \theta}}$
We can use this in the place of $\mathbf\small{|\vec{F_N}|}$ in (iv). We get:
$\mathbf\small{\frac{mg}{\cos \theta} \times \sin \theta=\frac{mv_0^2}{R}}$
$\mathbf\small{\Longrightarrow g \tan \theta=\frac{v_0^2}{R}}$
$\mathbf\small{\Longrightarrow {v_0^2}=Rg \tan \theta}$
Thus we get Eq.5.8:
$\mathbf\small{{v_0}=(Rg \tan \theta})^{\frac{1}{2}}$
• Note that, this result can be easily obtained by putting μ_s = 0 in Eq.5.7
• If a car move with this velocity on a banked road, friction at the contact surface will not be utilized. Thus wear and tear of the tires will be kept to a minimum.  

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Let us write a comparison of the two important results that we saw in the above discussion  
■ When the road is level, we have Eq.5.6: $\mathbf\small{v_{max}=\sqrt{\mu_s  Rg}}$ 
■ When the road is banked, we have Eq.5.7: $\mathbf\small{ {v_{max}}=\left[\frac{Rg(\mu_s+\tan \theta)}{1-\mu_s\tan \theta}\right]^{\frac{1}{2}}}$

• In both the equations there are three items on the right side

• In both the equations, all the three items come under 'square root'

• In both the equations, R and g are present. The difference is in the third item only

    ♦ In Eq.5.6, the third item is μ_s.

    ♦ In Eq.5.7, the third item is $\mathbf\small{\frac{(\mu_s+\tan \theta)}{1-\mu_s\tan \theta}}$

■ $\mathbf\small{\frac{(\mu_s+\tan \theta)}{1-\mu_s\tan \theta}}$ will be always greater than μ_s.

■ So we can write: When banking is provided, the car will be able to move with a greater speed

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In the next section, we will see some solved examples

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Chapter 5.21 - Rolling friction

In the previous section we saw coefficient of kinetic friction. We also saw some solved examples. In this section we will see a few more solved examples. After the solved examples we will see rolling friction.

Solved example 5.36
Two bodies A and B of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid wall (Fig. 5.82.a) The coefficient of friction between the bodies and the table is 0.15. A force of 200 N is applied horizontally to A. What are (a) the reaction of the partition (b) the action-reaction forces between A and B ? What happens when the wall is removed? Does the answer to (b) change, when the bodies are in motion? Ignore the difference between μ_s and μ_k.

Fig.5.82

Solution:
Case 1: When the wall is present
• The blocks will not be able to move even the 'smallest possible distance' because of the wall
■ The friction between the blocks and the ground will begin to act only when the blocks try to move. 
• This is because, force will not develop in the interlocking ridges and valleys (and also the adhesion) if the blocks do not move
• So in this case, we need not consider friction at all
Part (a):
1. The FBD of (A+B) is shown in fig.b
(The vertical forces will cancel each other and hence are not shown)
• A force of 200 N acts on the blocks from left to right
• Clearly, a 200 N force must act in the opposite direction. That is., from right to left. Other wise there will not be equilibrium
2. This 200 N from right to left is provided by the wall
• That means., the reaction from the wall is 200 N
Part (b):
1. Fig.c shows the FBD of A
• A force of 200 N acts on A from left to right
• Clearly, a 200 N force must act from right to left. Other wise there will not be equilibrium
2. This 200 N from right to left is provided by the block B
• That means., the reaction from B is 200 N
3. By the third law, this reaction from B must be due to the action from A
• Action and reaction are equal in magnitude and opposite in direction
• So we can write:
    ♦ A applies an action of 200 N on B (from left to right) 
    ♦ B applies a reaction of 200 N on A (from right to left)
Case 2: When the wall is absent:
• Now the blocks are free to move as shown in fig.5.83(a) below:

Fig.5.83

• We must first check whether the friction is strong enough to prevent motion:
• The FBD of (A+B) is shown in fig.5.83(b)
(The vertical forces will cancel each other and hence are not shown. But remember that, the vertical reaction from the surface is required for calculating the frictional force)

• We see that, the force from left to right is 200 N

• The force in the opposite direction (from right to left) is due to the frictional forces
• The total frictional force is: $\mathbf\small{\vec{f_{s,max(A)}}+\vec{f_{s,max(B)}}}$
• The magnitude of this total friction works out to:
$\mathbf\small{(\mu_s \times m_A \times g)+(\mu_s \times m_B \times g)=[\mu_s \times g \times (m_A+m_B)]=[0.15 \times 10 \times (5+10)]=22.5\,\text{N}}$
■ Since the 200 N is greater than this frictional force, the blocks will move
Now we can write the steps:
1. Since the two blocks move, the frictional force will be kinetic. So in the FBD, we must use $\mathbf\small{\vec{f_{k(A)}}\,\,\text{and}\,\,\vec{f_{k(B)}}}$
• This is shown in fig.5.83(c)
• When two forces act, the net force is given by the vector sum:
$\mathbf\small{\vec{F_1}+\vec{F_2}}$
2. $\mathbf\small{\text{Let}\,\,\vec{F_1}=200\,\text{N}}$
$\mathbf\small{\text{Let}\,\,\vec{F_2}=\vec{f_{k(A)}}+\vec{f_{k(B)}}}$
• The magnitude of this total friction works out to:
$\mathbf\small{(\mu_k \times m_A \times g)+(\mu_k \times m_B \times g)=[\mu_k \times g \times (m_A+m_B)]=[0.15 \times 10 \times (5+10)]=22.5\,\text{N}}$
• Note that, we are asked to ignore the difference between μ_s and μ_k. So we took μ_k = 0.15
• Thus we get: $\mathbf\small{\vec{F_2}=22.5\,\text{N}}$
• Considering forces towards right as positive and those towards left as negative, we get:
Net force on (A+B) = 200-22.5 = 177.5 N
3. By the second law, net force = mass × acceleration
• So we get: $\mathbf\small{177.5=m_{A+B} \times |\vec{a}|}$
$\mathbf\small{\Longrightarrow 177.5=15 \times |\vec{a}|}$
$\mathbf\small{\Longrightarrow |\vec{a}|=11.83\,\text{ms}^{-2}}$
• That means, the two blocks move together with an acceleration of 11.83 ms^-2.
4. The FBD of A is shown in fig.5.83(d) above
We see three forces acting on it. When three forces act, the net force is given by the vector sum:
$\mathbf\small{\vec{F_1}+\vec{F_2}+\vec{F_3}}$
5. $\mathbf\small{\text{Let}\,\,\vec{F_1}=200\,\text{N}}$
$\mathbf\small{\text{Let}\,\,\vec{F_2}=\vec{F_{N(AB)}}}$
$\mathbf\small{\text{Let}\,\,\vec{F_3}=\vec{f_{k(A)}}}$
• The magnitude of the frictional force $\mathbf\small{\vec{F_3}}$ works out to:
$\mathbf\small{(\mu_k \times m_A \times g)=(0.15 \times 5 \times 10)=7.5\,\text{N}}$
• Considering forces towards right as positive and those towards left as negative, we get:  
Net force on A = $\mathbf\small{200 \hat{i}-\vec{F_{N(AB)}}-7.5 \hat{i}}$
6. By the second law, net force = mass × acceleration
• So we get: $\mathbf\small{192.5 \hat{i}-\vec{F_{N(AB)}}=(m_{A} \times |\vec{a}|)\hat{i}}$
$\mathbf\small{\Longrightarrow 192.5\hat{i}-\vec{F_{N(AB)}}=(5 \times 11.83)\hat{i}}$
$\mathbf\small{\Longrightarrow \vec{F_{N(AB)}}=133.35\hat{i}}$
$\mathbf\small{\Longrightarrow |\vec{F_{N(AB)}}|=133.35\,\,\text{N}}$
7. By the third law, this reaction from B must be due to the action from A
• Action and reaction are equal in magnitude and opposite in direction
• So we can write:
    ♦ A applies an action of 133.35 N on B (from left to right) 
    ♦ B applies a reaction of 133.35 N on A (from right to left)

Solved example 5.37
The rear side of a truck is open and a box A of 40 kg mass is placed 5 m away from the open end as shown in Fig. 5.84 below. The coefficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 ms^-2. At what distance from the starting point does the box fall off the truck? (Ignore the size of the box)

Fig.5.84

Solution:
• First we have to determine whether the box will slip under the acceleration of 2 ms^-2
• We have: $\mathbf\small{|\vec{a_{max}}|=\mu_s \times g}$ (Details here)
• Substituting the values, we get: $\mathbf\small{|\vec{a_{max}}|=0.15 \times 10=1.5\,\text{ms}^{-2}}$
• But the truck accelerates at 2 ms^-2. So the box will definitely slip. That means, it will be sliding on the surface of the truck. 
Now we can write the steps:
1. If the surface of the truck is smooth, the box will not even move
• In that case, when the truck moves 5 m, the box will fall off
• Since such a situation does not arise, we can be sure that: Some force is dragging the box so that it moves with the truck
2. This dragging force is nothing but the frictional force between the truck and the box
• As the box is sliding, this frictional force is the 'kinetic friction'.
• We know how to calculate it's magnitude:
$\mathbf\small{|\vec{f_k}|=\mu_k \times |\vec{F_N}|=0.15 \times 400 = 60\, \text{N}}$
3. Now the FBD of the box A will be as shown in fig.5.85(a) below:

Fig.5.85

• We see that, 60 N is the only force acting on A
• So we can write: $\mathbf\small{m_A \times |\vec{a_A}|= 60\,\text{N}}$
$\mathbf\small{\Longrightarrow |\vec{a_A}|= \frac{60}{40}=1.5\,\text{ms}^{-2}}$
• That means, A moves towards left with an acceleration of 1.5 ms^-2
4. Remember that the truck is also moving towards the left. But it's acceleration is 2 ms^-2.
• So A will not be able to keep up with the truck. It will fall off after some time
• Before the motion begins, mark an arrow on the platform of the truck, right below the box. Let the tip of the arrow be 'P'
• This is shown in fig.5.85(b)
5. Let the truck start it's motion when the stop watch reading = 0
• Let the box fall off when stop watch reading = t seconds
• Let the box travel a distance of 'x' m during this 't' seconds
• Then, during this 't' seconds, point P must have traveled (5+x) meters
• This is shown in fig.c
6. So we have two sets of information:
(i) The box started from rest
• It moved with an acceleration of 1.5 ms^-2.
• It travelled for 't' s
• It travelled a distance of 'x' m during those 't' seconds 
(ii) The point 'P' started from rest
• It moved with an acceleration of 2.0 ms^-2.
• It travelled for 't' s
• It travelled a distance of '(5+x)' m during those 't' seconds 
7. We can use the equation: $\mathbf\small{s=ut+\frac{1}{2}at^2}$ 
• Applying the equation for the box, we get:
$\mathbf\small{x=0 \times t+\frac{1}{2}\times 1.5 \times t^2}$
$\mathbf\small{\Longrightarrow x=0.75 \times t^2}$
• Applying the equation for the point P, we get:
$\mathbf\small{5+x=0 \times t+\frac{1}{2}\times 2 \times t^2}$
$\mathbf\small{\Longrightarrow 5+x= t^2}$
• Solving the two equations, we get:
t = √20 s and x = 15 m
8. Thus we can write:
• When 'P' reaches a distance of (5+15) = 20 m from it's initial position, the box will fall off
• If 'P' travels 20 m, the truck also travels the same 20 m. 
■ So we can write:
When the truck reaches a distance of 20 m from it's initial position, the box will fall off

Solved example 5.38
In fig.5.86(a) below, a block 'A' of mass 4 kg is placed above another block 'B' of mass 5 kg. A force of 12 N (applied on 'A') is required to move 'A'. Then what is the maximum force that can be applied on 'B' so that 'A' and 'B' move together?

Fig.5.86

Solution:
1. Given that 12 N is the minimum required force. 
• That means, a force less than 12 N will not be sufficient to move A
• That means, 12 N is the limiting force $\mathbf\small{|\vec{f_{s,max}}|}$
2. We have:
$\mathbf\small{|\vec{f_{s,max}}|=\mu_{s,AB} \times |\vec{F_N}|}$
• Where μ_s,AB is the coefficient of static friction between A and B
• Substituting the values, we get: 12 × μ_s,AB = 40
⇒ μ_s,AB = 0.3
3. Next, we want to apply a force on B so that A and B 'move together'. 
• That means, A must not slip on B
• For small forces, A will not slip. But if the force is large, it will slip. 
• We want the maximum largest force which can be applied without causing the slip
4. We have: a_max = μ_s×g (Details here)
• Substituting the values, we get: a_max = 0.3 ×10 = 3 ms^-2
• That means., B can move with a maximum acceleration of 3 ms^-2
5. We have: Force = mass × acceleration
• But A and B are moving together. So mass is the total mass of (A+B) which is equal to 9 kg
• So we have a 9 kg mass moving at an acceleration of 3 ms^-2 
• So the required force = mass × acceleration = 9 × 3 = 27 N

Solved example 5.39
A block of mass 5 kg rests on an inclined plane as shown in fig.5.85(b) above. When the angle θ is 30^o, the block just starts sliding down. The coefficient of friction is 0.2. What is the velocity of the block 5 seconds after beginning the slide?
Solution:
• In this problem, the coefficient of friction is given as 0.2
• It is not specified whether 0.2 is μ_k or μ_s
• But we can confirm that it is μ_k 
• Because, μ_s will be tan 30, which is equal to 0.58
Now we can write the steps:
1. Resolving the forces parallel and perpendicular to the inclined plane, we get:
• Force causing the slide = mg sinθ
• Force resisting the slide = μ_k × mg cosθ
(See details here)
2. So net force = (mg sinθ - μ_k × mg cosθ) = mg(sinθ - μ_k cosθ) 
• Substituting the values, we get:
Net force = 5×10 (sin 30 - 0.2 × cos 30) = 16.34 N
3. Acceleration = ^{Net force}⁄_mass = ^16.34⁄₅  = 3.27 ms^-2.
4. We can use the equation: $\mathbf\small{v=u+at}$ 
• Substituting the values, we get:
$\mathbf\small{v=0 +3.27 \times 5=16.35\, \text{ms}^{-1}}$

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Rolling Friction

In fig.5.87(a) below, a wheel is rolling over a horizontal plane. 

Fig.5.87

• At any instant, there is only a ‘point of contact’ between the wheel and the plane. 
• It is just like a tangent drawn to a circle. We know that, a tangent to a circle will touch it only at one point.
• Remember that, in all the cases that we saw so far in this chapter, there is an ‘area of contact’. 
    ♦ But here, for the wheel, there is only a ‘point of contact’.
• Consider any such point at an instant. This point has no relative motion with the plane. 
    ♦ This is because, the next instant, another point will be in contact.
• So there should not be any static or kinetic friction between the wheel and the surface. 
• That means, once the wheel is set rolling, it should continue to roll even without any external force. 
• But in practice, we do not see this happening. 
The reason can be written in the following 4 steps:
1. The ‘point of contact’ between the wheel and the surface is an ‘ideal situation’. But we do not get such a situation in practice. 
2..This is because, a small deformation happens to both the wheel and the surface at the point of contact. 
3. As a result, there will indeed be a ‘small area of contact’. This results in friction. 
• The deformation if enlarged, will look as in fig.c.
4. The deformation is however ‘momentary’. That means, when the next portion of the wheel and surface come into contact, the earlier deformations will recover their original shapes

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• Thus we see that, friction comes into play even for the rolling motion. 

• But when compared to the static and kinetic friction, this rolling frictional force is very small in magnitude. 
• That means, if we can introduce rollers at the contact surface between two sliding objects, the resistance to sliding will be very low. 
• So ‘Invention of the wheel’ was indeed a major milestone in the development of mankind

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In the next section we will see circular motion

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Chapter 5.20 - The Coefficient of Kinetic Friction

In the previous section we saw coefficient of static friction. In this section we will see the kinetic friction.

• What happens if we apply a force greater than $\mathbf\small{|\vec{f_{s,max}}|}$?
• We already saw the answer in the previous section. We will write it again:

Behavior of the ridges and valleys:
• When the force exceeds $\mathbf\small{|\vec{f_{s,max}}|}$, the block will have to rise a little higher up so that, it’s inverted ridges gets freed from the valleys of the horizontal surface
• We do not notice this ‘rising of the block’ because, it is at a microscopic scale
• Once they are freed, motion can take place
• Also, when motion takes place, the tips of the ridges on both sides will be knocked off

Behavior of the adhesion
• When the force exceeds $\mathbf\small{|\vec{f_{s,max}}|}$, the ‘bonds of adhesion’ will break.
• Once those bonds are broken, motion can take place

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• So if the applied force is greater than $\mathbf\small{|\vec{f_{s,max}}|}$, the object will be in motion 
• During motion also, the object will experience friction
The frictional force experienced during motion is called Kinetic friction. It is denoted as: $\mathbf\small{\vec{f_k}}$  

• But the interlocking which takes place during motion will not be as effective as when the object is at rest
• Also new adhesive bonds will not be effectively formed when the object is in motion
■ In short, the object will be experiencing a lesser friction during motion. We can write:
$\mathbf\small{|\vec{f_s}|<|\vec{f_k}|}$
• We saw that, the static friction $\mathbf\small{|\vec{f_s}|}$ can reach a maximum value of $\mathbf\small{|\vec{f_{s,max}}|}$
• The kinetic friction (also known as sliding friction) does not have such a maximum value. 
• Even if the velocity of an object changes to a greater value, the magnitude of the frictional force $\mathbf\small{\vec{f_k}}$ opposing that motion remains the same

• Similarly, even if the velocity of an object changes to a lesser value, the magnitude of the frictional force $\mathbf\small{\vec{f_k}}$ opposing that motion remains the same

■Like static friction, kinetic friction also depends on the normal reaction $\mathbf\small{ |\vec{F_N}|}$. So, here also we have a similar relation:

5.3: $\mathbf\small{|\vec{f_k}|=\mu_s |\vec{F_N}|}$

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Now let us analyse an object in motion. We will write it in steps:

1. Fig.5.79(a) below shows a block of mass ‘m’ kg

Fig.5.79

• Under the influence of an external force $\mathbf\small{\vec{F}}$, it is moving with an acceleration $\mathbf\small{\vec{a}}$ 
• It is moving on a horizontal surface which is neither too smooth or too rough
2. The block is chosen as a sub-system as shown in fig.b
• In the FBD shown in fig.c, we see two forces
(i) $\mathbf\small{\vec{F}}$ towards right
(ii) $\mathbf\small{\vec{f_k}}$ towards left
• The resultant of the two forces is given by the vector sum: $\mathbf\small{\vec{F}-\vec{f_k}}$
3. By the second law, this resultant must be equal to m×a
• So we get: $\mathbf\small{\vec{F}-\vec{f_k}=m \times \vec{a}}$
■ From this we get the relation:
5.4: $\mathbf\small{\vec{a}=\frac{\vec{F}-\vec{f_k}}{m}}$
4. If the body is moving with a constant velocity, the acceleration will be zero
• So from the above relation, we get:
$\mathbf\small{0=\frac{\vec{F}-\vec{f_k}}{m}}$
$\mathbf\small{\Rightarrow \vec{F}=\vec{f_k}}$
We see an interesting fact here:
• If we see an object in uniform motion on an ordinary surface, it does not mean that no external force is acting on it. On ordinary surface, there need to be an external force even if the motion is uniform. This external force will be cancelled by the kinetic frictional force
• Earlier, we saw objects in uniform motion on frictionless surfaces. There is no need for an external force in such cases    
5. If we remove the external force, the relation becomes:
$\mathbf\small{\vec{a}=\frac{0-\vec{f_k}}{m}}$
$\mathbf\small{\Rightarrow \vec{a}=-\frac{\vec{f_k}}{m}}$
■That means, if the external force is removed, the object will begin to experience a negative acceleration
■All bodies which experience a negative acceleration will slow down and come to a stop

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Solved example 5.35
What is the acceleration of the block and trolley system shown in fig.5.80(a) below, if the coefficient of kinetic friction between the trolley and the surface is 0.04? What is the tension in the string? Assume the string to be light and inextensible [g = 10 ms^-2]

Fig.5.80

Solution:
1. The sub-systems are shown in fig.5.80.b
    ♦ Two arrows are shown at the ends of the string
    ♦ This is to help us remember that, a string in tension always pulls at it's ends (Details here)
• The FBD of ‘A’ is shown in fig.c
2. We see two forces acting on A. 
(The vertical forces will cancel each other and hence are not shown. But remember that, the vertical reaction from the surface is required for calculating the frictional force)
• When two forces act, the net force is given by the vector sum:
$\mathbf\small{\vec{F_1}+\vec{F_2}}$
3. $\mathbf\small{\text{Let}\,\,\vec{F_1}=\vec{T}\,\,\text{and}\,\,\vec{F_2}=\vec{f_k}}$
• Considering forces towards right as positive and those towards left as negative, we get:
Net force on A = $\mathbf\small{\vec{T}-\vec{f_k}}$
4. By the second law, net force = mass × acceleration
• So we get: $\mathbf\small{\vec{T}-\vec{f_k}=m_A \times \vec{a}}$ 
$\mathbf\small{\Rightarrow (|\vec{T}|)\hat{i}-(\mu_k \times |\vec{F_N}|)\hat{i}=(m_A \times |\vec{a}|)\hat{i}}$
$\mathbf\small{\Rightarrow (|\vec{T}|)\hat{i}-(\mu_k \times m_A \times g)\hat{i}=(m_A \times |\vec{a}|)\hat{i}}$
• Substituting the values, we get:
$\mathbf\small{(|\vec{T}|)\hat{i}-(0.04 \times 20 \times 10)\hat{i}=(20 \times |\vec{a}|)\hat{i}}$
$\mathbf\small{\Rightarrow |\vec{T}|-8=(20 \times |\vec{a}|)}$ 
5. The FBD of ‘B’ is shown in fig.d
We see two forces acting on B. When two forces act, the net force is given by the vector sum:
$\mathbf\small{\vec{F_1}+\vec{F_2}}$
6. $\mathbf\small{\text{Let}\,\,\vec{F_1}=\vec{T}\,\,\text{and}\,\,\vec{F_2}=\vec{W_B}}$
• Considering upward forces as positive and downward forces as negative, we get:
Net force on B = $\mathbf\small{\vec{T}-\vec{W_B}}$
7. By the second law, net force = mass × acceleration
• So we get: $\mathbf\small{\vec{T}-\vec{W_B}=-m_B \times \vec{a}}$ 
$\mathbf\small{\Rightarrow (|\vec{T}|)\hat{j}-(m_B \times g)\hat{j}=-(m_B \times |\vec{a}|)\hat{j}}$
• Substituting the values, we get:
$\mathbf\small{(|\vec{T}|)\hat{j}-(3 \times 10)\hat{j}=-(3 \times |\vec{a}|)\hat{j}}$
$\mathbf\small{\Rightarrow |\vec{T}|-30=-(3 \times |\vec{a}|)}$
8. Thus we get two equations:
• From (4) we have: $\mathbf\small{|\vec{T}|-8=(20 \times |\vec{a}|)}$
• From (7) we have: $\mathbf\small{|\vec{T}|-30=-(3 \times |\vec{a}|)}$
Solving them, we get: 
• $\mathbf\small{|\vec{a}|=\frac{22}{23}=0.96\,\text{ms}^{-2}}$
• $\mathbf\small{|\vec{T}|=27.1\,\text{N}}$

Problems in Limiting state of Static friction

• We have seen the basics of both static friction and kinetic friction. Now we can learn about the 'state' between the two types of frictions. 
• This 'state' is called the 'limiting state' because, if the external force is increased even by the smallest amount, the object will move. 
• We will write the analysis in steps:
1. Consider the long 'platform on wheels' shown in fig.5.81 below:

Fig.5.81

• The block 'A' of mass 'm' kg is resting on it
2. Initially, the platform is at rest. So the block is also at rest. 
• No horizontal forces are acting at this stage
3. Now, the platform starts to move. Some horizontal forces come into play. Let us see what they are:
• The platform starts from rest (velocity = 0) and attains a velocity v. 
• So surely, there has to be an acceleration
• Let the acceleration be done gradually. 
    ♦ That is., the initial acceleration $\mathbf\small{\vec{a_1}}$ is small 
• Then the block will move along with the platform with no slipping
4. However, the block will be trying to stay at it's position due to it's inertia
• But it cannot keep it's position
■ That means, some force is dragging the block
5. Obviously it is the frictional force between the block and the platform which causes the drag
• More precisely, it is the static friction between the block and the platform
• We can be sure that 'it is the static friction' and not 'the kinetic friction' because, there is no slipping between the block and the platform
• kinetic friction will come into play only when there is 'relative motion' between the block and the platform   
• Here, the platform is accelerating gradually
    ♦ The block is in motion relative to the ground
    ♦ But the block is stationary relative to the platform
• That is why we say: 'at this stage, there is no relative motion between the block and the platform'. 
• And so, the force of friction is that of 'static friction'
6. How does the static friction cause the drag?
Answer: It is through the interlocking and adhesion that we saw in fig.5.76 of the previous section
7. So the block is now moving as if it is clamped to the platform. 
• The 'clamping force' is the 'force of static friction' $\mathbf\small{\vec{f_s}}$ 
• But this force has an upper limit. We denoted it as $\mathbf\small{\vec{f_{s,max}}}$ 
8. It is our duty to ensure that $\mathbf\small{\vec{f_s}}$ do not reach $\mathbf\small{\vec{f_{s,max}}}$
■ If it does, any further slightest increase will cause the block to slip
• If we want to prevent 'some thing', we must know the 'cause'
• So in this case, we must know 'cause of increase in $\mathbf\small{\vec{f_s}}$'
9. Imagine that, the platform is now moving with a greater acceleration $\mathbf\small{\vec{a_2}}$
• This higher acceleration $\mathbf\small{\vec{a_2}}$ should be provided for the block also.
• Then only it can keep up with the platform
10. So a force of $\mathbf\small{m_A \times \vec{a_2}}$ will be felt by the block 
• The medium through which the acceleration is provided to the block is the same interlocking and adhesion that we saw earlier in fig.5.76 of the previous section
11. But due to inertia, the block does not want to receive this new acceleration. It wants to stay back
• However, due to the interlocking, the acceleration will be transfered 
• So the ridges and valleys (and also the adhesive bonds) will begin to 'feel' the higher force $\mathbf\small{m_A \times \vec{a_2}}$
12. But there is a 'limiting force' which those ridges and valleys (and also the adhesive bonds) can take.
• We know that ' the limiting force' is the $\mathbf\small{\vec{f_{s,max}}}$
13. So if the $\mathbf\small{m_A \times \vec{a_2}}$ exceeds $\mathbf\small{\vec{f_{s,max}}}$, the ridges and valleys (and also the adhesive bonds) will break. The block will slip
• We must see that $\mathbf\small{m_A \times \vec{a}}$ is kept low. We cannot decrease m_A. So the only option is to keep the acceleration low
14. The maximum acceleration possible can be obtained by equating the two forces. So we can write:
$\mathbf\small{m_A \times \vec{a_{max}}=\vec{f_{s,max}}}$
$\mathbf\small{\Longrightarrow m_A \times |\vec{a_{max}}|=|\vec{f_{s,max}}|}$
$\mathbf\small{\Longrightarrow m_A \times |\vec{a_{max}}|=\mu_s \times |\vec{F_N}|}$
$\mathbf\small{\Longrightarrow m_A \times |\vec{a_{max}}|=\mu_s \times m_A \times g}$
$\mathbf\small{\Longrightarrow |\vec{a_{max}}|=\mu_s \times g}$
• We see that, the 'maximum acceleration possible' is independent of the mass

An example:
Determine the maximum acceleration of a train in which a box lying on it's floor, will remain stationary, given that the coefficient of static friction between the box and the floor of the train is 0.15
Solution:
We have: $\mathbf\small{|\vec{a_{max}}|=\mu_s \times g}$  
Substituting the values, we get: $\mathbf\small{|\vec{a_{max}}|=0.15 \times 10=1.5\,\text{ms}^{-2}}$

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In the next section we will see a few more solved examples

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