Question

the diagram shows the forces acting on a block resting on an inclined plane. the angle θ is adjusted until the block is just at the point of sliding. r is the normal reaction, w the weight of the block and f the maximum frictional force. not to scale what is the maximum coefficient of static friction between the block and the plane? a sinθ b cosθ c tanθ d 1/tanθ

Explanation:

Step1: Resolve weight into components

Resolve the weight $W$ of the block into components parallel and perpendicular to the inclined - plane. The component of the weight parallel to the inclined plane is $W\sin\theta$ and the component perpendicular to the inclined plane is $W\cos\theta$.

Step2: Analyze forces at equilibrium

When the block is just at the point of sliding, the maximum frictional force $F$ is equal to the component of the weight parallel to the inclined plane, i.e., $F = W\sin\theta$, and the normal reaction $R=W\cos\theta$.

Step3: Recall friction formula

The formula for the maximum static - frictional force is $F=\mu_sR$, where $\mu_s$ is the coefficient of static friction.

Step4: Substitute and solve for $\mu_s$

Substitute $F = W\sin\theta$ and $R = W\cos\theta$ into $F=\mu_sR$. We get $W\sin\theta=\mu_sW\cos\theta$. Canceling out $W$ from both sides of the equation, we have $\mu_s=\frac{\sin\theta}{\cos\theta}=\tan\theta$.

Answer:

C. $\tan\theta$