More Solved examples involving Torque and Center of gravity

This page shows two more solved examples in continuation of the examples that we saw here: Solved examples involving torque.

Example 1
From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body
Solution:
1. The uniform circular disc can be supported at it's center as shown in fig.1(a) below:

Fig.1

• But when a hole is cut on the right side, the disc will tilt towards the left and fall off. This is shown in fig.1(b)
2. If we shift the support towards the left (along the diameter), the disc will balance again
• This is shown in fig.2(a) below:

Fig.2

• In fig.2(a), the disc is supported at the new CG
• We have to find the position of the new CG
3. In fig.2(b), the small circle is painted red and glued back to get the original disc
• The CG of this 'composite disc' will be same as the center of the original disc 
4. In such a situation, we apply the conditions of equilibrium
• Let the brown portion in fig.2(b) be indicated as 'Brown'
• Let the red portion in fig.2(b) be indicated as 'Red'
• Let 'mass per unit area' of the disc be M
• Since the disc is uniform, this 'M' will be same at all points on the disc
5. Fig.3(a) below shows the dimensions

Fig.3

• Area of Red = $\mathbf\small{\pi\,\left(\frac{R}{2}\right)^2=\frac{\pi R^2}{4}}$
    ♦ So $\mathbf\small{W_{Red}=\frac{\pi R^2Mg}{4}}$
• Area of Brown = $\mathbf\small{\pi\,R^2-\frac{\pi R^2}{4}=\frac{3\pi R^2}{4}}$
    ♦ So $\mathbf\small{W_{Brown}=\frac{3\pi R^2Mg}{4}}$ 
6. Fig.3(b) shows the forces
• The distance between the required CG and the support is denoted as 'x'
7. Since the composite disc is in translational equilibrium, we have:
R - W_Brown - W_Red = 0
8. Since the composite disc is in rotational equilibrium, we have: net torque = 0
• Let us take the torques about the support 
(i) Torque created by W_Brown about the support = W_Brown × x (anti clockwise)
(ii) Torque created by R about the support = 0
(iii) Torque created by W_Red about the support = W_Red × ^R⁄₂ (clockwise)
9. So applying the condition, we get:
-(W_Brown × x) + (W_Red × ^R⁄₂) = 0
$\mathbf\small{\Rightarrow x=\frac{R\,W_{Red}}{2\,W_{Brown} }}$
• Substituting the known values, we get: $\mathbf\small{x=\frac{\pi R^3Mg}{4}\div \frac{6\pi R^2Mg}{4}}$
$\mathbf\small{\Rightarrow x=\frac{\pi R^3Mg}{4}\times \frac{4}{6\pi R^2Mg}=\frac{R}{6}}$

Example 2
A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?
Solution:
1. Given that, the metre stick is balanced on a knife edge at its centre 
• That means, if the support is given at the exact center, the metre stick will balance
• That means, it is a uniform scale
• This is shown in fig.2(a) below:

Fig.2

2. The condition when the two coins are placed at the 12 cm mark is shown in fig.b
• The forces and dimensions are also shown
• We see that, for equilibrium, the support has to be shifted by 5 cm towards the left
3. Since the system in fig.b is in translational equilibrium, we have:
R - W_Coin - W_Stick = 0
8. Since the system is in rotational equilibrium, we have: net torque = 0
• Let us take the torques about the support 
(i) Torque created by W_Coin about the support = W_Coin × 0.33 (anti clockwise)
(ii) Torque created by R about the support = 0
(iii) Torque created by W_Stick about the support = W_Stick × 0.05 (clockwise)
9. So applying the condition, we get:
-(W_Coin × 0.33) + (W_Stick × 0.05) = 0
W_Stick = (W_Coin × ^0.33⁄_0.05 ) = (10 ×  ¹⁄₁₀₀₀ ×  ^0.33⁄_0.05 ) = 0.066 kg = 66 g

---

PREVIOUS           CONTENTS

Copyright©2019 Higher Secondary Physics. blogspot.in - All Rights Reserved

Chapter 7.22 - Solved examples involving Torque

In the previous section, we saw center of gravity. In this section we will see some solved examples

Solved example 7.21
The metal bar in fig.7.103(a) is 0.70 m long and has a mass of 4 kg. It is supported on two knife edges placed 0.10 m from each end. A 6 kg mass is suspended from a point P, which is 0.20 m from the left support

Fig.7.103

Find the reactions at the supports. Assume the bar is of uniform cross section and homogeneous. Take g = 9.8 ms^-2
Solution:
1. Let us name the supports as A and B
■ The weight of the rod acts at it’s CG
• Given that, the metal bar is of uniform cross section and homogeneous.
■ So the CG of the rod will be at it’s geometric center
2. The detailed measurements are shown in fig.b
The CG is marked as G
3. Since the bar is in translational equilibrium, we have:
R_A + R_B - 6g - 4g = 0
⇒ R_A + R_B = 10g = 98.0 N.
4. Since the bar is in rotational equilibrium, we have: net torque = 0
• Let us take the torques about the support A
(i) Torque created by R_A about A = zero
(ii) Torque created by 6g about A = 6g × 0.2 = 1.2g Nm (clockwise)
(iii) Torque created by 4g about A = 4g × 0.25 = 1.0g Nm (clockwise)
(iv) Torque created by R_B about A = R_B × 0.5 Nm (anti clockwise)
5. So applying the condition, we get:
1.2g + g - 0.5R_B = 0
⇒ R_B = 43.12 N
• Substituting this value of R_B in (3), we get: R_A = (98-43.12) = 54.88 N

Solved example 7.22
A 3 m long ladder having a mass of 20 kg, leans on a frictionless wall. It’s feet rest on the ground 1 m from the wall as shown in fig.7.104(a) below:

Fig.7.104

Find the reaction forces on the wall and the floor. Take g = 9.8 ms^-2
Solution:
1. Let the end points of the ladder be A and B
• A is 1 m from the wall. This is shown in fig.b
• Let C be the foot of the wall
2. The reaction from the floor at A will be normal to the floor
• This reaction is denoted as R_A
3. The frictional force prevents the point A from moving away from C
• This frictional force is denoted as F. It pulls the ladder towards C. Other wise the ladder will slip
• Also recall that, the frictional force is always parallel to the surface
4. The reaction from the wall at B will be normal to the wall
• This reaction is denoted as R_B
5. Given that, the wall is frictionless
• If there was friction, a force F would have acted parallel to the wall in the upward direction
• This force would have made some contribution towards: 'preventing the movement of B towards C'   • In other words, this force would have made some contribution towards: 'preventing the ladder from slipping'
• In addition to that, this force would have made some contribution towards resisting the vertical force (20 g) of the ladder
    ♦ Where 'g' is the acceleration due to gravity
• But in this problem there is no such force
    ♦ The slipping is prevented entirely by the horizontal force F at A
    ♦ The vertical load 20 g is resisted entirely by the vertical force R_A at A
6. The weight of the ladder is 20 g
• It acts downwards at the CG of the ladder
• The CG is marked as G in fig.b
7. The above steps gives us all the 4 forces acting on the ladder
• Now we can apply the conditions of equilibrium
• Since the ladder is in translational equilibrium, we have:
(i) In the x direction: F - R_B = 0
(ii) In the y direction: R_A - 20 g = 0
⇒ R_A = 20 g = 196 N
8. Since the bar is in rotational equilibrium, we have: net torque = 0
Let us take the torques about A
(i) Torque created by R_A about A = zero
(ii) Torque created by F about A = zero
(iii) Torque created by R_B about A = R_B × BC
So we want the length of BC
• Applying Pythagoras theorem to the right triangle ABC, we get:
$\mathbf\small{BC=\sqrt{AB^2-AC^2}=\sqrt{3^2-1^2}=2\sqrt{2}\;\text{m}}$
Thus the required torque = $\mathbf\small{2\sqrt{2}R_B\;\text{N m}}$ (clockwise) 
(iv) Torque created by 20 g about A = 20 g × AD
• So we want the length of AD
• D is the foot of the perpendicular drawn from G
• Consider the similar triangles ABC and AGD
• We have: $\mathbf\small{\frac{AD}{AC}=\frac{AG}{AB}}$
$\mathbf\small{\Rightarrow \frac{AD}{1}=\frac{1.5}{3}}$
⇒ AD = 0.5 m
• Thus the torque = 20 g × 0.5 × 9.8 = 98 N m (anti clockwise)
9. Applying the condition, we get:
$\mathbf\small{2\sqrt{2}R_B-98=0}$
⇒ R_B = 34.65 N
10. Substituting this value of R_B in 7(i), we get: F = 34.65 N.
11. At the point A, two forces are acting on the ladder:
• R_A vertically and F horizontally. They are shown in fig.c
• The resultant of the two forces = $\mathbf\small{\sqrt{(R_A)^2+F^2}=\sqrt{196^2+34.65^2}=}$ 199.04 N
• Let this resultant make an angle α with the horizontal
• Then $\mathbf\small{\alpha=\tan^{-1}\frac{R_A}{F}=\tan^{-1}\frac{196}{34.65}=}$ 79.97^o

Solved example 7.23 
A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in fig.7.105(a) below:

Fig.7.105

The angles made by the strings with the vertical are 36.9^o and 53.1^o respectively. The bar is 2 m long. Calculate the distance d of the center of gravity of the bar from it’s left end
Solution:
1. The free body diagram is shown in fig.b
• Let T₁ and T₂ be the tensions in the strings
• In fig.a, we see that, the left string makes an angle of 36.9^o with the vertical wall
    ♦ In fig.b, we see that, this string makes the same angle with the vertical dotted line
    ♦ The angles are same because, they are alternate angles 
• In fig.a, we see that, the right string makes an angle of 53.1^o with the vertical wall
    ♦ In fig.b, we see that, this string makes the same angle with the vertical dotted line
    ♦ The angles are same because, they are alternate angles
• Now we can resolve the tensions into horizontal and vertical components. This is shown in fig.c
2. Since the bar is in translational equilibrium, we have:
(i) In the x direction: - T₁ sin θ₁ + T₂ sin θ₂  = 0
⇒ T₁ sin θ₁ = T₂ sin θ₂.
⇒ T₁ sin 36.9 = T₂ sin 53.1 = T₂ cos (90 - 36.9) = T₂ cos 36.9
⇒ $\mathbf\small{\frac{T_2}{T_1}=\tan36.9 =0.75}$
⇒ T₂ = 0.75 T₁
(ii) In the y direction: T₁ cos θ₁ + T₂ cos θ₂ - W = 0
⇒ T₁ cos 36.9 + T₂ cos 53.1 = W
⇒0.8 T₁ + 0.6 T₂ = W
3. Since the bar is in rotational equilibrium, we have: net torque = 0
• Let us take the torques about G
(i) Torque created by T₁ sin θ₁ about G = zero
(ii) Torque created by T₁ cos θ₁ about G = T₁ cos θ₁ × d (clockwise)
(iii) Torque created by W about G = zero
(iv) Torque created by T₂ cos θ₂ about G = T₂ cos θ₂ × (2-d) (anti clockwise)
(i) Torque created by T₂ sin θ₂ about G = zero
4. Applying the condition, we get:
[T₁ cos θ₁ × d] - [T₂ cos θ₂ × (2-d)] = 0
⇒ [T₁ × cos 36.9 × d] - [0.75 T₁ × cos 53.1 × (2-d)] = 0
(∵ from 2(i), we have: T₂ = 0.75 T₁)
⇒ [0.8 T₁ d] - [0.45 T₁ (2-d)] = 0
⇒ 0.8 T₁ d - 0.9 T₁ + 0.45 T₁ d = 0
⇒ 1.25 T₁ d = 0.9 T₁
⇒ 1.25 d = 0.9
⇒ d = 0.72 m

Solved example 7.24
A car weighs 1800 kg. The distance between it’s front and back axles is 1.8 m. It’s center of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel
Solution:
1. Fig.7.106 below shows the schematic diagram

Fig.106

• The small grey circles at the center of the wheels denote the axles
• The front and rear axles are named as A and B respectively
2. When the car is in translational equilibrium, we have:
R_A + R_B -1800g = 0
3. When the car is in rotational equilibrium, we have: net torque = 0
Let us take the torques about G
(i) Torque created by R_A about G = 1.05 R_A (clockwise) 
(ii) Torque created by 1800g about G = zero
(iii) Torque created by R_B about G = 0.75 R_B (anticlockwise)
4. Applying the condition, we get:
1.05 R_A - 0.75 R_B = 0
1.05 R_A = 0.75 R_B
R_B = 1.4 R_A
5. Substituting this in (2), we get:
R_A + 1.4R_A = 1800 × 9.8
⇒ R_A = 7350 N
• So R_B = 1.4 × 7350 = 10290 N
6. The reaction on the front axle A is 7350 N
• But the load from this axle is resisted by two front wheels
• So reaction on each of the front wheels = ⁷³⁵⁰⁄₂ = 3675 N
7. The reaction on the back axle B is 10290 N
• But the load from this axle is resisted by two back wheels
• So reaction on each of the back wheels = ¹⁰²⁹⁰⁄₂ = 5145 N

Solved example 7.25
As shown in fig.7.107(a), the two sides of a step ladder BA and CA are 1.6 m long and hinged at A. A rope DE 0.5 m long is tied half way up. A weight 40 kg is suspended from a point F, 1.2 m from B along the ladder BA. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. Take g = 9.8 ms^-2 [Hint: Consider the equilibrium of each side of the ladder separately]

Fig.7.107

Solution:
1. The detailed measurements are shown in fig.b
• Given:
    ♦ BD = 0.8 m
    ♦ DF = FA = 0.4 m
    ♦ DE = 0.5 m
• Let ∠ABC = θ
    ♦ Since DE is parallel to the ground, we get: ∠ADE = θ   
2. Let us calculate the other required dimensions:
• A perpendicular is dropped from A onto BC
    ♦ This is shown as red dashed line 
    ♦ The foot of the perpendicular on BC is O
    ♦ The foot of the perpendicular on DE is G
• Applying Pythagoras theorem to the right triangle ADG, we get:
$\mathbf\small{AG=\sqrt{AD^2-DG^2}=\sqrt{0.8^2-0.25^2}=0.76\;\text{m}}$
3. In the similar triangles ABO and ADG, we get:
• $\mathbf\small{\frac{AD}{AB}=\frac{AG}{AO}}$
$\mathbf\small{\Rightarrow \frac{0.8}{1.6}=\frac{0.76}{AO}}$
⇒ AO = 1.52 m
4. Again in the same similar triangles ABO and ADG, we get:
$\mathbf\small{\frac{AD}{AB}=\frac{DG}{BO}}$
$\mathbf\small{\Rightarrow \frac{0.8}{1.6}=\frac{0.25}{BO}}$
⇒ BO = 0.5 m
5. A perpendicular is dropped from F onto AG
• The foot of this perpendicular is H
• In the similar triangles AFH and ADG, we get:
$\mathbf\small{\frac{AF}{AD}=\frac{FH}{DG}}$
$\mathbf\small{\Rightarrow \frac{0.4}{0.8}=\frac{FH}{0.25}}$
⇒ FH = 0.125 m
6. Let us write the above distances together:
    ♦ BD = 0.8 m
    ♦ DF = FA = 0.4 m
    ♦ DE = 0.5 m
    ♦ AG =0.76 m
    ♦ AO = 1.52 m
    ♦ BO = 0.5 m
    ♦ FH = 0.125 m
• Thus we obtained all the distances that we will soon require for torque calculations
7. The forces are shown in fig.c below:

Fig.7.107 (c) & (d)

• R_B is the normal reaction at B
• R_C is the normal reaction at C
• Note that the tensions T in the string are internal forces and so will cancel each other
8. For translational equilibrium, we have:
• R_B + R_C - (40 × 9.8) = 0
⇒ R_B + R_C - 392 = 0
⇒ R_B + R_C = 392
9. For rotational equilibrium, we need to calculate the torques first:
• Let us find the torques about O
(i) Torque created by R_B about O = R_B × BO = R_B × 0.5 = 0.5R_B Nm (clockwise)
(ii) Torque created by the 40 kg load about O = (40 × 9.8) × FH = 392 0.125 = 49 Nm (anti clockwise)
(iii) Torque created by R_C about O = R_C × CO = R_C × 0.5 = 0.5R_C Nm (anti clockwise)
10. Applying the condition, we get:
⇒ 0.5R_B - 49 - 0.5R_C = 0
⇒ 0.5R_B - 0.5R_C = 49
⇒ R_B - R_C = 98
11. Solving the equations in (8) and (10), we get:
R_B = 245 N, R_C = 147 N
12. Next we want to find T
• In the fig.d above, the FBD of side AB is shown separately
• The forces acting are: R_B, T and the load of 40 kg
• Let us calculate the torques about A:
(i) Torque created by R_B about A = R_B × BO = R_B × 0.5 = 0.5R_B Nm (clockwise)
(ii) Torque created by T about A = T × AG = T × 0.76 = 0.76T Nm (anti clockwise)
(ii) Torque created by the 40 kg load about A = (40 × 9.8) × FH = 392 × 0.125 = 49 Nm (anti clockwise)
13. Applying the condition, we get:
0.5R_B - 0.76T - 49 = 0
0.5 × 245 - 0.76T - 49 = 0
T = 96.7 N

---

Two more solved examples can be seen here

---

In the next section, we will see moment of inertia

PREVIOUS           CONTENTS          NEXT

Copyright©2019 Higher Secondary Physics. blogspot.in - All Rights Reserved

Chapter 7.20 - Principle of Moments

In the previous section, we saw conditions of equilibrium of rigid bodies. In this section we will see Principle of Moments

1. In the fig.7.97(a) below, a light (ie. of negligible mass) rod AB lies on the xy-plane
• It’s length is (d₁+d₂)

Fig.7.97

• It is supported at point 'C' by a 'knife edge support'
• The 'knife edge support' is shown clearly in the 3D view in fig.7.98 below:

Fig.7.98

2. Two forces are applied on the rod
• Those two forces have the following five peculiarities:
(i) The forces do not have the same magnitude
(ii) Both forces have the same directions
    ♦ Let us assume the usual orientation: x-axis is horizontal and y-axis is vertical
    ♦ Then both forces act towards the -ve side of the y-axis
(iii) One force is applied at A
(iv) The other force is applied at B
(v) Both forces are perpendicular to AB
3. As a result of the forces $\mathbf\small{\vec{F}_A}$ and $\mathbf\small{\vec{F}_B}$, a reaction force will develop at C
• This reaction force is denoted as $\mathbf\small{\vec{R}_C}$
• $\mathbf\small{\vec{R}_C}$ is shown to be vertical and directed towards the +ve side of the y-axis
• We will soon prove mathematically that, it is indeed vertical and directed towards the +ve side of y-axis
4. We want the rod AB to remain in equilibrium (both translational and rotational)
• How can we achieve that?
Ans: By ensuring that, the conditions of equilibrium are satisfied
• So let us apply the conditions:
• We have already seen that, if all the forces lie in the xy-plane, we need to check 3 conditions only:
$\mathbf\small{\sum{\vec{F}_{x,ext}}=\vec{0}}$
$\mathbf\small{\sum{\vec{F}_{y,ext}}=\vec{0}}$
$\mathbf\small{\sum{\vec{\tau}_{z,ext}}=\vec{0}}$
• Further, there are no forces in the x-direction. So we need to check 2 conditions only:
$\mathbf\small{\sum{\vec{F}_{y,ext}}=\vec{0}}$
$\mathbf\small{\sum{\vec{\tau}_{z,ext}}=\vec{0}}$
5. First we will apply:
$\mathbf\small{\sum{\vec{F}_{y,ext}}=\vec{0}}$
• If this condition is to be satisfied, the vector sum ($\mathbf\small{\vec{F}_A+\vec{F}_B+\vec{R}_C}$) must be a zero vector
• This condition tells us that, if there is to be translational equilibrium, $\mathbf\small{\vec{R}_C}$ is indeed vertical
• The reason can be written in 5 steps:
(i) If $\mathbf\small{\vec{R}_C}$ is inclined, it will have a vertical as well as a horizontal component
(ii) But we see no other horizontal components to counter it
(iii) So, if there is to be translational equilibrium, $\mathbf\small{\vec{R}_C}$ must be truly vertical
(iv) Also, it must be directed towards the +ve side of the y-axis
(v) Only then we will get ($\mathbf\small{\vec{F}_A+\vec{F}_B+\vec{R}_C}$) equal to a zero vector
6. Since we know the directions, we can write the above equation in terms of unit vectors also:
$\mathbf\small{-|\vec{F}_A|\hat{j}-|\vec{F}_B|\hat{j}+|\vec{R}_C|\hat{j}=\vec{0}}$
$\mathbf\small{\Rightarrow \left(-|\vec{F}_A|-|\vec{F}_B|+|\vec{R}_C|\right)\hat{j}=\vec{0}}$
• In the above equation, we have vectors on both sides. So their magnitudes must be equal
• Thus the condition becomes:
$\mathbf\small{-|\vec{F}_A|-|\vec{F}_B|+|\vec{R}_C|=0}$

---

7. For applying the next condition, we have to first find the magnitudes and directions of the torques created by the forces in the xy-plane
• Let ‘C’ be the origin. Then we can mark the position vectors as shown in fig.b
• Torque created by $\mathbf\small{\vec{F}_A}$ about C 
= $\mathbf\small{(\vec{r}_A \times \vec{F}_A)}$
• Magnitude of this torque
= $\mathbf\small{|\vec{r}_A|\times |\vec{F}_A|\times \sin \theta}$
= $\mathbf\small{[d_1\times |\vec{F}_A|\times \sin 90]=d_1\times |\vec{F}_A|}$
(θ = 90^o because, forces are perpendicular to the rod)
8. For finding the direction of this torque:
(i) Assume that $\mathbf\small{\vec{F}_A}$ is shifted so that, it's tail end coincides with the tail end of $\mathbf\small{\vec{r}_A}$  
(ii) Imagine that a right handed screw is placed perpendicular to the computer screen
(iii) Turn it from $\mathbf\small{\vec{r}_A}$ to $\mathbf\small{\vec{F}_A}$
(iv) We see that, the screw moves away from the screen, towards us
(v) The rod is lying on the xy-plane. So the computer screen is the xy-plane
(vi) The z-axis is perpendicular to the computer screen
(vii) Let us assume that, the positive direction of the z-axis points towards us
(viii) So we can write:
The torque created by $\mathbf\small{\vec{F}_A}$ is perpendicular to the xy-plane and is directed towards the +ve side of the z axis
• Combining the magnitude and the direction, we can write:
The torque created by $\mathbf\small{\vec{F}_A}$ is: $\mathbf\small{[d_1\times |\vec{F}_A|]\; \hat{k}}$ 
9. Next we find the torque created by $\mathbf\small{\vec{F}_B}$ about C 
= $\mathbf\small{(\vec{r}_B \times \vec{F}_B)}$
• Magnitude of this torque
= $\mathbf\small{|\vec{r}_B|\times |\vec{F}_B|\times \sin \theta}$
= $\mathbf\small{[d_2\times |\vec{F}_B|\times \sin 90]=d_2\times |\vec{F}_B|}$
(θ = 90^o because, forces are perpendicular to the rod)
10. For finding the direction of this torque:
(i) Assume that $\mathbf\small{\vec{F}_B}$ is shifted so that, it's tail end coincides with the tail end of $\mathbf\small{\vec{r}_B}$  
(ii) Imagine that a right handed screw is placed perpendicular to the computer screen
(iii) Turn it from $\mathbf\small{\vec{r}_B}$ to $\mathbf\small{\vec{F}_B}$
(iv) We see that, the screw moves into the screen, away from us
(v) The rod is lying on the xy-plane. So the computer screen is the xy-plane
(vi) The z-axis is perpendicular to the computer screen
(vii) We already assumed that, the positive direction of the z-axis points towards us
(viii) So we can write:
The torque created by $\mathbf\small{\vec{F}_B}$ is perpendicular to the xy-plane and is directed towards the -ve side of the z axis
• Combining the magnitude and the direction, we can write:
The torque created by $\mathbf\small{\vec{F}_B}$ is: $\mathbf\small{[-d_2\times |\vec{F}_B|]\; \hat{k}}$ 
11. Next we find the torque created by $\mathbf\small{\vec{R}_C}$ about C
• But $\mathbf\small{\vec{R}_C}$ passes through C
• So the torque is $\mathbf\small{\vec{0}}$  
12. Now we can write a summary about the torques:
• The torque created by $\mathbf\small{\vec{F}_A}$:
$\mathbf\small{[d_1\times |\vec{F}_A|]\; \hat{k}}$
• The torque created by $\mathbf\small{\vec{F}_B}$:
$\mathbf\small{[-d_2\times |\vec{F}_B|]\; \hat{k}}$
• The torque created by $\mathbf\small{\vec{R}_C}$ is $\mathbf\small{\vec{0}}$
13. We want $\mathbf\small{\sum{\vec{\tau}_{z,ext}}=0}$
• Let us input the values:
$\mathbf\small{[(d_1\times |\vec{F}_A|)\hat{k}]-[(d_2\times |\vec{F}_B|)\hat{k}]+\vec{0}}$ must be equal to $\mathbf\small{\vec{0}}$
• That means, the following equation must be satisfied:
$\mathbf\small{[(d_1\times |\vec{F}_A|)-(d_2\times |\vec{F}_B|)]\hat{k}=\vec{0}}$
14. In the above equation, we have vectors on both sides. So their magnitudes must be equal
• Thus the condition becomes:
$\mathbf\small{[(d_1\times |\vec{F}_A|)-(d_2\times |\vec{F}_B|)]=0}$
15. The above equation can be rearranged as: $\mathbf\small{|\vec{F}_B|=\frac{|\vec{F}_A|d_1}{d_2}}$
• From the fig.7.97, we see that d₁ is less than d₂. So $\mathbf\small{\frac{d_1}{d_2}}$ will be less than '1'
• Thus we find that: $\mathbf\small{\vec{F}_B}$ is less than $\mathbf\small{\vec{F}_A}$
16. Also we can write:
• The magnitude of the force applied at B should be equal to $\mathbf\small{\frac{|\vec{F}_A|d_1}{d_2}}$
• Then only we can achieve equilibrium for the rod AB
17. We can write the other way also:
• The magnitude of the force applied at A should be equal to $\mathbf\small{\frac{|\vec{F}_B|d_2}{d_1}}$
• Then only we can achieve equilibrium for the rod AB

---

Let us see a practical application of the equation:

In fig.7.99 below, a block of 10 kg hangs from the end A of the rod AB. When the rod is exactly horizontal, the block rests completely on the floor and the string is straight.

Fig.7.99

What force must be applied at B so that, the block just loses contact with the floor? 
[Take g = 10 ms^-2]
Solution:
1. We have: $\mathbf\small{\vec{F}_A}$ = mg = 10 × 10 = 100 N
• Using equation in (16) above, we get:
$\mathbf\small{|\vec{F}_B|=\frac{|\vec{F}_A|d_1}{d_2}=\frac{100 \times 0.2}{0.8}=25\, \text{N}}$
2. 25 N corresponds to a mass of 2.5 kg
• So we are able to lift a large mass of 10 kg by a small mass of 2.5 kg

---

Note that, the 25 N corresponds to 'just losing contact between the block and the floor'

• If we lift the block further, the rod AB will no longer be horizontal
• As a result, the forces $\mathbf\small{\vec{F}_A}$ and $\mathbf\small{\vec{F}_B}$ will no longer be perpendicular to AB
• This situation is shown in fig.7.99(b)
■ If we want to keep the block stationary at a height much above the floor, what force must be applied at B?
Solution:
1. Fig.7.100(a) below shows a schematic diagram

Fig.7.100

AB makes an angle θ with the horizontal
2. We see that:
• $\mathbf\small{\vec{F}_A}$ and $\mathbf\small{\vec{F}_B}$ are vertical
    ♦ But $\mathbf\small{\vec{F}_A}$ is not perpendicular to AB
    ♦ $\mathbf\small{\vec{F}_B}$ is also not perpendicular to AB
3. So we resolve the two forces into components:
(i) $\mathbf\small{\vec{F}_A}$ has two components:
• One component is parallel to AB. It has magnitude: $\mathbf\small{|\vec{F}_A|\sin \theta}$ 
• The other component is perpendicular to AB. It has magnitude: $\mathbf\small{|\vec{F}_A|\cos \theta}$
(ii) $\mathbf\small{\vec{F}_B}$ has two components:
• One component is parallel to AB. It has magnitude: $\mathbf\small{|\vec{F}_B|\sin \theta}$ 
• The other component is perpendicular to AB. It has magnitude: $\mathbf\small{|\vec{F}_B|\cos \theta}$
• These components are shown in blue color in fig.b
4. The parallel components pass through C
• So they do not create any torque about C
• We need to consider the perpendicular components only
5. We can easily show the following two items:
(i) The torque created by the perpendicular component of $\mathbf\small{\vec{F}_A}$ is:
$\mathbf\small{[d_1\times |\vec{F}_A|\times \cos \theta]\; \hat{k}}$
(ii) The torque created by the perpendicular component of $\mathbf\small{\vec{F}_B}$ is:
$\mathbf\small{[-d_2\times |\vec{F}_B|\times \cos \theta]\; \hat{k}}$
(The reader may write the proofs in his/her own note books)
6. We want $\mathbf\small{\sum{\vec{\tau}_{z,ext}}=0}$
• Let us input the values:
$\mathbf\small{[(d_1\times |\vec{F}_A|\times \cos \theta)\hat{k}]-[(d_2\times |\vec{F}_B|\times \cos \theta)\hat{k}]}$ must be equal to $\mathbf\small{\vec{0}}$
• That means, the following equation must be satisfied:
$\mathbf\small{\left[(d_1\times |\vec{F}_A|\times \cos \theta)-(d_2\times |\vec{F}_B|\times \cos \theta)\right]\hat{k}=\vec{0}}$
14. In the above equation, we have vectors on both sides. So their magnitudes must be equal
• Thus the condition becomes:
$\mathbf\small{\left[(d_1\times |\vec{F}_A|\times \cos \theta)-(d_2\times |\vec{F}_B|\times \cos \theta)\right]=0}$
$\mathbf\small{\Rightarrow (d_1\times |\vec{F}_A|)-(d_2\times |\vec{F}_B|)=0}$
• This is the same result that we obtained earlier
15. The above equation can be rearranged as: $\mathbf\small{|\vec{F}_B|=\frac{|\vec{F}_A|d_1}{d_2}}$
16. Let us input the values:
$\mathbf\small{|\vec{F}_B|=\frac{100 \times 0.2}{0.8}=25\, \text{N}}$
• This is the same result that we obtained before

---

■ The rod AB in fig.7.99 is a lever

Let us see the important features of a lever:
(i) An ideal lever is a light (ie. of negligible mass) rod pivoted at a point along its length
(ii) The pivot point is called the fulcrum
(iii) The force $\mathbf\small{\vec{F}_A}$ is usually some weight to be lifted
• This $\mathbf\small{\vec{F}_A}$ is called the load
• Distance (d₁) of load from the fulcrum is called load arm
(iv) The force $\mathbf\small{\vec{F}_B}$ applied to lift the load
• This $\mathbf\small{\vec{F}_B}$ is called the effort
• Distance (d₂) of effort from the fulcrum is called effort arm
(v) Consider the expression '$\mathbf\small{(d_1\times |\vec{F}_A|)-(d_2\times |\vec{F}_B|)=0}$' that we derived in (14) above
• It can be rearranged as: $\mathbf\small{d_1\times |\vec{F}_A|=d_2\times |\vec{F}_B|}$
• That is: Load arm × Load = Effort arm × Effort
• This is the mathematical expression of the Principle of moments for a lever
(vi) The ratio $\mathbf\small{\frac{Load}{Effort}}$ is called Mechanical advantage
• From (v) above, we get: mechanical advantage = $\mathbf\small{\frac{\text{Effort arm}}{\text{Load arm}}}$
(vii) If $\mathbf\small{\frac{\text{Effort arm}}{\text{Load arm}}}$ is greater than '1', we will be able to lift a larger load by a smaller effort
(viii) A see saw at the children’s park is an example of a lever
Some other examples are:
Cutting pliers
Wheel barrow
Nut cracker etc.

---

In the next section, we will see Center of gravity

PREVIOUS           CONTENTS          NEXT

Copyright©2019 Higher Secondary Physics. blogspot.in - All Rights Reserved