Question

Two blocks of equal weights are connected by a string as shown. If the blocks slide with constant velocity, find the coefficient of kinetic friction ($\mu_k$).

$30^\circ$

Answer 1

The coefficient of kinetic friction is $2-\sqrt{3}$.

Answer 2

Let $W$ be the weight of each block. Since the blocks slide with constant velocity, the net force on each block is zero.

Case 1: Block on the incline moves down the incline. Let Block 1 be on the incline and Block 2 be on the horizontal. For Block 1 (on incline): The component of weight down the incline is $W \sin(30^\circ)$. The tension $T$ acts up the incline. The kinetic friction $f_{k1}$ acts up the incline, opposing motion. The normal force $N_1 = W \cos(30^\circ)$. The kinetic friction $f_{k1} = \mu_k N_1 = \mu_k W \cos(30^\circ)$. For constant velocity, the net force parallel to the incline is zero: $W \sin(30^\circ) - T - f_{k1} = 0$ $W \sin(30^\circ) - T - \mu_k W \cos(30^\circ) = 0$ (Equation 1)

For Block 2 (on horizontal): The tension $T$ pulls to the right. The kinetic friction $f_{k2}$ acts to the left, opposing motion. The normal force $N_2 = W$. The kinetic friction $f_{k2} = \mu_k N_2 = \mu_k W$. For constant velocity, the net force is zero: $T - f_{k2} = 0$ $T - \mu_k W = 0$ (Equation 2)

From Equation 2, $T = \mu_k W$. Substitute this into Equation 1: $W \sin(30^\circ) - \mu_k W - \mu_k W \cos(30^\circ) = 0$ Divide by $W$: $\sin(30^\circ) - \mu_k - \mu_k \cos(30^\circ) = 0$ $\sin(30^\circ) = \mu_k (1 + \cos(30^\circ))$ $\mu_k = \frac{\sin(30^\circ)}{1 + \cos(30^\circ)}$ Given $\sin(30^\circ) = \frac{1}{2}$ and $\cos(30^\circ) = \frac{\sqrt{3}}{2}$: $\mu_k = \frac{1/2}{1 + \sqrt{3}/2} = \frac{1/2}{(2+\sqrt{3})/2} = \frac{1}{2+\sqrt{3}}$ Rationalizing the denominator: $\mu_k = \frac{1}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}} = \frac{2-\sqrt{3}}{4-3} = 2-\sqrt{3}$.

Case 2: Block on the incline moves up the incline. If Block 1 moves up the incline, then Block 2 moves to the left. For Block 1: $T - W \sin(30^\circ) - \mu_k W \cos(30^\circ) = 0$ For Block 2: $T - \mu_k W = 0$ This would lead to $\mu_k = 2+\sqrt{3}$, which is an unrealistically high value for kinetic friction. Therefore, the motion is down the incline.