Question

the drawings show identical blocks at rest on three different inclines, each block held in place by static friction. rank the static friction in the three cases in ascending order (smallest first).

Explanation:

Step1: Analyze force equilibrium

For a block at rest on an incline of angle $\theta$ with mass $m$, the force of static - friction $f_s$ balances the component of the gravitational force along the incline $mg\sin\theta$. So $f_s = mg\sin\theta$.

Step2: Compare angles

The steeper the incline, the larger the angle $\theta$. Since the blocks are identical (same $m$), and $g$ is constant, the force of static friction is directly proportional to $\sin\theta$. In the given inclines, the angle of incline $B$ is the smallest, followed by $C$, and then $A$ (where $A$ has the steepest incline). So, $\sin\theta_B<\sin\theta_C<\sin\theta_A$, and thus $f_{sB}

Answer:

B, C, A