Chapter 5.13 - Normal reaction on Inclined surfaces

In the previous section we saw cases in which:
    ♦ Contact surfaces were all vertical
    ♦ So  the $\mathbf\small{\vec{F_N}}$ were all horizontal
In the section just before that, we saw cases in which:
    ♦ Contact surfaces were all horizontal. 
    ♦ So the $\mathbf\small{\vec{F_N}}$ were all vertical. 
In this section we will see cases in which:
    ♦ Contact surfaces are inclined

1. Consider Fig.5.41(a) below. A wooden block (B) is resting on a wedge shaped block (I). 
• The top surface of this wedge is a smooth inclined plane. 

Fig.5.41

• The plane is inclined at an angle θ to the horizontal
■ Before learning about the interaction between the block and the wedge, we must learn some geometrical features of the wedge. Steps (2) to (4) below enable us to learn those features:
2. The top surface of the wedge is not horizontal. It is inclined
• Draw a perpendicular to this inclined surface. It is the white dashed line in fig.5.41(b)
    ♦ This perpendicular intersects the inclined surface at D
    ♦ This perpendicular intersects the base of the wedge at C
• Draw a vertical through D. It is the green dashed line in fig.b
    ♦ This vertical intersects the base of the wedge at B
3. The base of the wedge is horizontal
• So now we have two right triangles:
    ♦ ⊿ACD right angled at D
    ♦ ⊿BCD right angled at B
4. In ⊿ACD, we have ∠ACD = (90-θ)
• From this we get: In ⊿BCD, ∠BDC = θ
■ This is an important result. It tells us that:
The following two angles are always the same:
(i) The angle between ‘the inclined surface’ and the 'horizontal'
    ♦ In the fig., it is the ∠DAB
(ii) The angle between ‘the normal to the inclined surface’ and the 'vertical'
    ♦ In the fig., it is the ∠CDB
5. Now we can learn about the interaction
• The $\mathbf\small{\vec{W_B}}$ in fig.a, denotes the weight of the block
• It will be always directed verticaly downwards (towards the center of the earth)   
• It's direction will not depend on the inclination of the plane
3. We can resolve $\mathbf\small{\vec{W_B}}$ into it's two perpendicular components
• For the convenience of resolution, let us choose the frame of reference (coordinate axes) in this way (see fig.5.41c):
    ♦ The x axis is parallel to the inclined surface
    ♦ The y axis is perpendicular to the inclined surface
4. So we have a new set of coordinate axes
• Consider fig.d. It shows a vector $\mathbf\small{\vec{F}}$ in the usual set of coordinate axes
• In this case, we know that:
    ♦ x component = $\mathbf\small{\vec{F_x}=(|\vec{F}|\cos \theta)\hat{i} }$ 
    ♦ y component = $\mathbf\small{\vec{F_y}=(|\vec{F}|\sin \theta)\hat{j} }$
(Details about the above results can be seen here)
• So we see that:
    ♦ The component which is adjacent to the angle gets the cosine
    ♦ The other component gets the sine
5. Now consider fig.e. It shows or present case
• Here θ is adjacent to the y axis 
Clearly:    
    ♦ x component = $\mathbf\small{\vec{W_{Bx}}=(|\vec{W_{B}}|\sin \theta)\hat{i} }$
    ♦ y component = $\mathbf\small{\vec{W_{By}}=(|\vec{W_{B}}|\cos \theta)\hat{j} }$
• The two components are shown in fig.5.42(a) below:

Fig.5.42

6. Next step is to draw the FBD of the block
• The component of $\mathbf\small{\vec{W_B}}$ parallel to the x axis is $\mathbf\small{\vec{W_{Bx}}}$ 
    ♦ This force is exerted by the earth. It is a force acting on the block. 
    ♦ So we show it in the FBD
• The component of $\mathbf\small{\vec{W_B}}$ parallel to the y axis is $\mathbf\small{\vec{W_{By}}}$ 
    ♦ This force is exerted by the earth. It is a force acting on the block. 
    ♦ So we show it also in the FBD
• Due to the action $\mathbf\small{\vec{W_{By}}}$ on the inclined surface, the inclined surface produces a reaction. 
• We can denote it as $\mathbf\small{\vec{F_{N(BI)}}}$
    ♦ This $\mathbf\small{\vec{F_{N}}}$ acts on the block. 
    ♦ So we show it in the FBD
• The FBD is now complete. It is shown in the fig.5.42(b)
7. We observe no movement in the y direction
• That is., there is equilibrium in the y direction
• So we can write: $\mathbf\small{\vec{W_{By}}=-\vec{F_{N(BI)}}}$
8. Now we consider the x direction
• There is no force to oppose $\mathbf\small{\vec{W_{Bx}}}$ 
(Remember that, the surface in this case is smooth. So there is no friction)
• So the block will move in the direction of $\mathbf\small{\vec{W_{Bx}}}$
• That means, the block will slide down along the inclined surface
• If there is friction between the ‘bottom surface of the block’ and the 'inclined surface', the block may be able to hold it’s position. This happens when the frictional force is equal in magnitude to $\mathbf\small{\vec{W_{Bx}}}$. We will learn more about  friction in later sections of this chapter

Solved example 5.21
A block of mass 3 kg slides down along a friction less inclined surface. What is the acceleration with which it slides down? Angle of inclination of the surface with the horizontal is 30^o. Take g = 10 ms^-2.
Solution:
1. Let the x axis be parallel to the inclined plane
• Let the y axis be perpendicular to the inclined plane
2. The force which causes the sliding is: 
$\mathbf\small{\vec{W_{Bx}}=(|\vec{W_{B}}|\sin \theta)\hat{i} }$
• Substituting the values, we get:
$\mathbf\small{\vec{W_{Bx}}=(mg \,\sin 30)\hat{i} }$
• So the magnitude of the force which causes the sliding is: mg sin30 = 0.5mg
3. Let 'a' be the acceleration with which the block slides
• Then we have: ma = 0.5mg
⟹ a = 0.5g = 0.5 × 10 = 5 ms^-2
■ Note that, this acceleration does not depend on the mass. But it certainly depends on the 'angle which the inclined surface makes with the horizontal'

Solved example 5.22
Three blocks A, B and C are placed on the smooth inclined surface of a wedge as shown in fig.5.43(a) below:

Fig.5.43

The masses of the blocks A, B and C are 1 kg, 2 kg and 3 kg respectively. External forces of 60 N and 18 N also act on the blocks as shown.  
(a) Find the acceleration of A, B and C
(b) Draw the following FBDs:
(i) FBD of (A+B+C)
(ii) FBD of A
(iii) FBD of B
(iv) FBD of C

Solution:
1. FBD of (A+B+C) is shown in fig.5.43(b)
• The modified frame of reference is also shown
• There is no motion in the y direction. So we need not consider the forces in the that direction. They will cancel each other
2. In the x direction, we have:
Net force = $\mathbf\small{60\, \hat{i}-\vec{W_{(A+B+C)x}}-18\, \hat{i}}$
⟹ Net force = $\mathbf\small{42\, \hat{i}-\vec{W_{(A+B+C)x}}}$
⟹ Net force = $\mathbf\small{42\, \hat{i}-(|\vec{W_{(A+B+C)}|\, \sin \theta )\, \hat{i}}}$
⟹ Net force = $\mathbf\small{42\, \hat{i}-[(m_{(A+B+C)} \times g) \, \sin \theta ]\, \hat{i}}$
• Substituting the values, we get:
• Net force = $\mathbf\small{\left[ 42\, \hat{i}-[(6 \times 10) \, \sin 30 ]\, \hat{i} \right ]=12\, \hat{i}}$
4. Since there is a net force, there will be acceleration
• The net force is the cause of the acceleration of (A+B+C). Applying the second law, we can write:
$\mathbf\small{(m_{(A+B+C)}\times|\vec{a_x}|)\, \hat{i}=12\, \hat{i}}$
$\mathbf\small{\Rightarrow (m_{(A+B+C)}\times|\vec{a_x}|)=12}$
• Substituting the values, we get:
$\mathbf\small{(6\times|\vec{a_x}|)=12}$
$\mathbf\small{\Rightarrow |\vec{a_x}|=2\, \text{ms}^{-1}}$
■ So each of the three blocks move down with an acceleration of 2 ms^-2

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1. Now we draw the FBD of A. This is shown in the fig.5.44(a) below:

Fig.5.44

• There is no motion in the y direction. So we need not consider the forces in the that direction. They will cancel each other

2. In the x direction, we have:

Net force = $\mathbf\small{\vec{F_{N(AB)}}-\vec{W_{Ax}}-18\, \hat{i}}$

⟹ Net force = $\mathbf\small{\vec{F_{N(AB)}}-(|\vec{W_{A}}|\, \sin \theta )\, \hat{i}-18\, \hat{i}}$

⟹ Net force = $\mathbf\small{\vec{F_{N(AB)}}-(m_A \times g \times\sin \theta )\, \hat{i}-18\, \hat{i}}$

Substituting the values, we get:

Net force = $\mathbf\small{\vec{F_{N(AB)}}-(1 \times 10 \times\sin 30 )\, \hat{i}-18\, \hat{i}}$

⟹ Net force = $\mathbf\small{\vec{F_{N(AB)}}-5\, \hat{i}-18\, \hat{i}}$

⟹ Net force = $\mathbf\small{\vec{F_{N(AB)}}-23\, \hat{i}}$

3. This net force causes the acceleration of A. So we can write:

$\mathbf\small{\vec{F_{N(AB)}}-23\, \hat{i}=m_A \times \vec{a}}$

$\mathbf\small{\Rightarrow \vec{F_{N(AB)}}-23\, \hat{i}=(m_A \times |\vec{a}|)\, \hat{i}}$

$\mathbf\small{\Rightarrow \vec{F_{N(AB)}}-23\, \hat{i}=(1 \times 2)\, \hat{i}}$

$\mathbf\small{\Rightarrow \vec{F_{N(AB)}}= 25\, \hat{i}}$

$\mathbf\small{\Rightarrow |\vec{F_{N(AB)}}|= 25\, \text{N}}$

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1. Next we draw the FBD of B. This is shown in the fig.5.44(b) above

• There is no motion in the y direction. So we need not consider the forces in the that direction. They will cancel each other

2. In the x direction, we have:

Net force = $\mathbf\small{\vec{F_{N(BC)}}-\vec{W_{Bx}}-\vec{F_{N(BA)}}}$

⟹ Net force = $\mathbf\small{\vec{F_{N(BC)}}-(|\vec{W_{B}}|\, \sin \theta )\, \hat{i}-25\, \hat{i}}$

($\mathbf\small{\because\:  |\vec{F_{N(BA)}}|=|\vec{F_{N(AB)}}|= 25\, \text{N}}$)

⟹ Net force = $\mathbf\small{\vec{F_{N(BC)}}-(m_B \times g \times\sin \theta )\, \hat{i}-25\, \hat{i}}$

Substituting the values, we get:

Net force = $\mathbf\small{\vec{F_{N(BC)}}-(2 \times 10 \times\sin \text{30} )\, \hat{i}-25\, \hat{i}}$

⟹ Net force = $\mathbf\small{\vec{F_{N(BC)}}-10\, \hat{i}-25\, \hat{i}}$

⟹ Net force = $\mathbf\small{\vec{F_{N(BC)}}-35\, \hat{i}}$

3. This net force causes the acceleration of B. So we can write:

$\mathbf\small{\vec{F_{N(BC)}}-35\, \hat{i}=m_B \times \vec{a}}$

$\mathbf\small{\Rightarrow \vec{F_{N(BC)}}-35\, \hat{i}=(m_B \times |\vec{a}|)\, \hat{i}}$

$\mathbf\small{\Rightarrow \vec{F_{N(BC)}}-35\, \hat{i}=(2 \times 2)\, \hat{i}}$

$\mathbf\small{\Rightarrow \vec{F_{N(BC)}}= 39\, \hat{i}}$

$\mathbf\small{\Rightarrow |\vec{F_{N(BC)}}|= 39\, \text{N}}$

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1. Finally, we draw the FBD of C. This is shown in the fig.5.44(c) above

• There is no motion in the y direction. So we need not consider the forces in the that direction. They will cancel each other

2. In the x direction, we have:

Net force = $\mathbf\small{60 \, \hat{i}-\vec{W_{Cx}}-\vec{F_{N(CB)}}}$

⟹ Net force = $\mathbf\small{60 \, \hat{i}-(|\vec{W_{C}}|\, \sin \theta )\, \hat{i}-39\, \hat{i}}$

($\mathbf\small{\because\:  |\vec{F_{N(CB)}}|=|\vec{F_{N(BC)}}|= 39\, \text{N}}$)

⟹ Net force = $\mathbf\small{60 \, \hat{i}-(m_B \times g \times\sin \theta )\, \hat{i}-39\, \hat{i}}$ 

Substituting the values, we get:

Net force = $\mathbf\small{60 \, \hat{i}-(3 \times 10 \times\sin \text{30})\, \hat{i}-39\, \hat{i}}$  

⟹ Net force = $\mathbf\small{60 \, \hat{i}-15\, \hat{i}-39\, \hat{i}=6 \, \hat{i}}$

3. This net force causes the acceleration of C. So we can write:

$\mathbf\small{6 \hat{i}=m_C \times \vec{a}}$

$\mathbf\small{\Rightarrow 6 \hat{i}=(m_C \times |\vec{a}|)\, \hat{i}}$

$\mathbf\small{\Rightarrow 6 \hat{i}=(3 \times |\vec{a}|)\, \hat{i}}$
$\mathbf\small{\Rightarrow |\vec{a}|=2\, \text{ms}^{-1}}$
This is indeed true because, we saw that all the 3 blocks move with the same acceleration of 2 ms^-2.

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In the next section we will see force in strings.

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