JEE Challenger
More from Laws of Motion
PhyHardAdv 2010 • Paper 1

Frictional Force vs Applied Force Graph on Inclined Plane

A block of mass mm is on an inclined plane of angle θ\theta. The coefficient of friction between the block and the plane is μ\mu and tan⁡θ>μ\tan\theta > \mu. The block is held stationary by applying a force PP parallel to the plane. The direction of force pointing up the plane is taken to be positive. As PP is varied from P1=mg(sin⁡θ−μcos⁡θ)P_1 = mg(\sin\theta - \mu \cos\theta) to P2=mg(sin⁡θ+μcos⁡θ)P_2 = mg(\sin\theta + \mu \cos\theta), the frictional force ff versus PP graph will look like

Diagram 0

Options

A
Option A Diagram 1
Correct
B
Option B Diagram 1
C
Option C Diagram 1
D
Option D Diagram 1

Topics & Concepts

Laws of MotionFriction

Step-by-Step Solution

To find the relationship between the frictional force ff and the applied force PP, let us analyze the forces acting on the block along the inclined plane.

Let the direction pointing up the inclined plane be the positive xx-direction. The forces acting along the incline are:

  1. The component of gravity acting down the incline: −mgsin⁡θ-mg\sin\theta
  2. The applied external force acting up the incline: +P+P
  3. The static frictional force: ff (taken along the incline with sign indicating its direction)

Since the block remains stationary throughout the range of applied force from P1P_1 to P2P_2, it is in static equilibrium. The net force acting along the inclined plane must be zero:
∑Fx=0\sum F_x = 0
P−mgsin⁡θ+f=0P - mg\sin\theta + f = 0

Solving for the frictional force ff:
f=mgsin⁡θ−Pf = mg\sin\theta - P

This equation represents a linear relationship between ff and PP with a constant negative slope of −1-1.

Evaluating ff at the boundaries:

  1. At the minimum applied force, P=P1=mg(sin⁡θ−μcos⁡θ)P = P_1 = mg(\sin\theta - \mu\cos\theta):
    f(P1)=mgsin⁡θ−mg(sin⁡θ−μcos⁡θ)=+μmgcos⁡θ>0f(P_1) = mg\sin\theta - mg(\sin\theta - \mu\cos\theta) = +\mu mg\cos\theta > 0
    The frictional force is positive (directed up the incline) and reaches its maximum static value to prevent the block from slipping downward.

  2. At P=mgsin⁡θP = mg\sin\theta:
    f=mgsin⁡θ−mgsin⁡θ=0f = mg\sin\theta - mg\sin\theta = 0
    The applied force exactly balances the component of gravity, so no friction is required.

  3. At the maximum applied force, P=P2=mg(sin⁡θ+μcos⁡θ)P = P_2 = mg(\sin\theta + \mu\cos\theta):
    f(P2)=mgsin⁡θ−mg(sin⁡θ+μcos⁡θ)=−μmgcos⁡θ<0f(P_2) = mg\sin\theta - mg(\sin\theta + \mu\cos\theta) = -\mu mg\cos\theta < 0
    The frictional force is negative (directed down the incline) and reaches its maximum static value to prevent the block from slipping upward.

Thus, as PP increases from P1P_1 to P2P_2, the frictional force ff decreases linearly from a positive value +μmgcos⁡θ+\mu mg\cos\theta, passes through zero, and ends at a negative value −μmgcos⁡θ-\mu mg\cos\theta.

This corresponds to the straight line with a negative slope shown in Option A.

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