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Question: An inclined plane is inclined at an angle \(\theta\) with the horizontal. A body of mass \(m\) rests...

An inclined plane is inclined at an angle θ\theta with the horizontal. A body of mass mm rests on it, if the coefficient of friction is μ\mu , then the minimum force that has to be applied to the inclined plane to make the body just most up the inclined plane is-
A. μmgcosθ\mu mg\cos \theta
B. mgsinθmg\sin\theta
C. μmgcosθmgsinθ\mu mg\cos \theta- mg\sin \theta
D. μmgcosθ+mgsinθ\mu mg\cos \theta + mg\sin \theta

Explanation

Solution

Draw the free body diagram of the given question by assuming the inclined plane as a triangle with the angle of θ\theta. Mark the xx and yy components of the body.

Step by Step Solution:
When a body is inclined with a plane at an angle of θ\theta its force can be distributed into two different components. Now, the free body diagram can be drawn as given below,

 ![](https://www.vedantu.com/question-sets/e2eff192-a0c1-4ff9-b5da-522d0e2fed5e2420846200146585174.png)  

In the above given figure,
μ\mu is the coefficient of friction, NN is the force acting, mm is the mass of the inclined body, gg is the acceleration due to gravity andθ\theta is the angle of inclination.
Now, the force acting due to the body of mass mm is mgmg. Dissipating it into the xx and yy components, we get mgsinθmg\sin \theta and mgcosθmg\cos \theta respectively. The force acting in the opposite direction in order to pull the mass just up to the inclined plane would be μN\mu N.
Now, from the figure A, we can see that
N=mgcosθN = mg\cos \theta , since they both are in the opposite direction
Therefore, in order to make the body just most up the inclined plane, a force greater than mgsinθmg\sin \theta should be applied,
Therefore,
Fmin=μNmgsinθ{F_ {\min}} = \mu N - mg\sin \theta
Where Fmin{F_ {\min}} is the minimum force required to move the body just up the inclined plane.
We know that
N=mgcosθN = mg\cos \theta
Therefore,
Fmin=μmgcosθmgsinθ{F_ {\min}} = \mu mg\cos \theta - mg\sin \theta
So, the net force along the inclined plane is μmgcosθmgsinθ\mu mg\cos \theta - mg\sin \theta
Therefore,
The minimum force required to move the body just up the inclined plane is μmgcosθmgsinθ\mu mg\cos \theta - mg\sin \theta
Or option C.

Note: The minimum force required to push the body up the incline plane. It must be equal to the net maximum force acting along the plane of incline. The friction for which the body does not move is called the static friction. The friction for which the body moves is called the kinetic friction. The minimum force required will be equal to maximum value of friction.