Question

A body of mass $ 2kg $ is lying on a rough inclined plane of inclination $ {30^ \circ } $ . Find the magnitude of the force parallel to the incline needed to make the block move 

(a) up the incline 

(b) down the incline. 

Coefficient of friction $ = 0.2 $ .

Answer 1

Hint: To solve this question, we have to find out the magnitude and the direction of the frictional force which acts on the block in both the cases. Then from the equilibrium of the block in the direction along the inclined plane, we will get values of both the forces.

Formula used: The formula used to solve this question is given by

 $ f = \mu N $ , here $ f $ is the force of friction acting between two surfaces where the coefficient of friction is $ \mu $ and the normal reaction is $ N $ .

Complete answer
Consider the block and the inclined plane as shown in the figure below.

We resolve the forces along and normal to the inclined plane. The block does not move perpendicular to the inclined plane. Therefore applying equilibrium along the direction normal to the inclined plane, we get

$ N - Mg\cos {{\theta = 0}} $

$ \Rightarrow N = Mg\cos {{\theta }} $ …………………...(1)

Now, the direction of the friction acting on the block will be opposite to the motion of the block.

(a) Let $ {F_U} $ be the force parallel to the incline needed to make the block move up the incline. As the block moves up the incline, so the friction will act down the incline, as shown below.

As the force $ {F_U} $ is just sufficient to make the block move up the inclined plane, so we have

 $ {F_U} = Mg\sin {{\theta }} + f $

Now, the force of friction is given by

 $ f = \mu N $

Substituting this above, we get

 $ {F_U} = Mg\sin {{\theta }} + \mu N $

From (1)

 $ {F_U} = Mg\sin {{\theta }} + \mu Mg\cos {{\theta }} $

 $ \Rightarrow {F_U} = Mg\left( {\sin {{\theta }} + \mu \cos {{\theta }}} \right) $

According to the question, $ M = 2kg $ , $ {{\theta }} = {30^ \circ } $ , and $ \mu = 0.2 $ . Also $ g = 10m/{s^2} $ . Substituting these above we get

 $ {F_U} = 2 \times 10\left( {0.5 + 0.2 \times 0.86} \right) $

 $ {F_U} = 13.44N $

Hence, the force needed to make the block move up the incline is equal to $ 13.44N $

(b) Let $ {F_D} $ be the force parallel to the incline needed to make the block move down the incline. As the block moves down the incline, so the friction will act up the incline, as shown below.

As the force $ {F_D} $ is just sufficient to make the block move down the inclined plane, so we have

$ {F_D} + Mg\sin {{\theta }} = f $

Now the force of friction is given by

$ f = \mu N $

First we have to check if the downward component of weight of block is sufficient for motion of block in downward direction. 

According to the question, $ M = 2kg $ , $ {{\theta }} = {30^ \circ } $ , and $ \mu = 0.2 $

$=Mg\sin\theta = 2 \times 9.8 \times \sin ( {30^ \circ }) $

$=9.8 N $

And Now Friction on block

$=\mu Mg\cos\theta = 0.2 \times 2 \times 9.8 \times \cos ( {30^ \circ }) $

$=0.2 \times 9.8 \times \sqrt{3} N $

$= 3.39 N$

So friction is already less than sine component of weight so the block will move without any external effort.

Substituting this above, we get

$ {F_D} =0N $

Hence, the force needed to make the block move down the incline is equal to $ 0N$.

Note:

The block is not in equilibrium along the inclined plane. Since the values of the evaluated forces are just sufficient to make the block move on the incline, we have applied equilibrium to get the minimum value of the force.