This problem involves analyzing the frictional force acting on a block held stationary on an inclined plane by an external force P. The key is to understand how the frictional force (f) changes as the applied force (P) is varied. Let's break it down step by step.

Step 1: Understand the Forces

The forces acting on the block parallel to the inclined plane are:

• The component of gravitational force down the plane: $mg\sin \theta$
• The applied force P (which can be positive up the plane or negative down the plane).
• The frictional force f, which opposes the tendency of motion.

Since the block is held stationary, the net force parallel to the plane is zero.

Step 2: Direction of Friction

The direction of friction depends on the value of P relative to the gravitational component.

• If P is less than $mg\sin \theta$, the block has a tendency to slide down. Hence, friction acts up the plane.
• If P is greater than $mg\sin \theta$, the block has a tendency to move up. Hence, friction acts down the plane.

Step 3: Equation for Equilibrium

For the block to be stationary, the sum of forces parallel to the plane is zero. We consider two cases based on the direction of friction:

Case 1: Friction acting up the plane (f positive)
This happens when P is relatively small. The equilibrium equation is: $P+f=mg\sin \theta$
So, $f=mg\sin \theta -P$

Here, f is static friction and can vary from 0 to its maximum value $\mu N=\mu mg\cos \theta$. This case is valid as long as $f\le \mu mg\cos \theta$ and P is between $mg(\sin \theta -\mu \cos \theta )$ and $mg\sin \theta$.

Case 2: Friction acting down the plane (f negative)
This happens when P is relatively large. The equilibrium equation is: $P=mg\sin \theta +f$
So, $f=P-mg\sin \theta$

Here, f is negative (since it acts down the plane) and its magnitude can vary from 0 to $\mu mg\cos \theta$. This case is valid for P between $mg\sin \theta$ and $mg(\sin \theta +\mu \cos \theta )$.

Step 4: Analyze the f vs P graph

From the equations:

• For $P\le mg\sin \theta$: $f=mg\sin \theta -P$. This is a straight line with slope -1 and intercept $mg\sin \theta$ on the f-axis.
• For $P\ge mg\sin \theta$: $f=P-mg\sin \theta$. This is a straight line with slope +1 and intercept $-mg\sin \theta$ on the f-axis.

However, the friction has a limit: $\left|f\right|\le \mu mg\cos \theta$. So the graph is V-shaped but with the tips cut off at the maximum friction values. The minimum P for which the block can be held stationary is $P1=mg(\sin \theta -\mu \cos \theta )$ (friction at its maximum up the plane) and the maximum P is $P2=mg(\sin \theta +\mu \cos \theta )$ (friction at its maximum down the plane). At $P=mg\sin \theta$, the friction is zero.

Final Answer: The f vs P graph is a V-shaped graph with its vertex at (P = mg sinθ, f = 0). The arms of the V have slopes of -1 and +1, but they are truncated at f = ±μmg cosθ. Therefore, the correct graph is the one that looks like an inverted 'V' with horizontal lines at the top and bottom, representing the maximum static friction limits.

Related Topics and Formulae

Friction on an Inclined Plane: The maximum static friction force is given by $f\max =\mu N=\mu mg\cos \theta$, where N is the normal force.

Equilibrium Condition: For an object to be stationary, the net force in any direction must be zero. Here, we resolved forces parallel to the inclined plane.

Graph Interpretation: The f vs P graph is piecewise linear with slopes determined by the equilibrium equations, and it is bounded by the maximum friction value.