GSO001 Problem 9

First, determine whether object B slides relative to board A.

Assuming A and B move together as one unit, let the common acceleration be $a_0$ . The equation of motion for the system:

(M + m)a_0 = F

The only force accelerating object B is the static friction force $f$ from A:

m a_0 = f

For B not to slide on A, this static friction $f$ must not exceed the maximum static friction $\mu m g$ :

f \le \mu m g \implies m a_0 \le \mu m g \implies a_0 \le \mu g

The maximum force $F_0$ for A and B to move together:

F_0 = (M + m) \mu g

Substituting numerical values:

F_0 = (3.0 + 2.0) \times 0.50 \times 9.8 = 24.5 \ \text{N}

Since the applied force $F = 30 \ \text{N} > F_0$ , the maximum static friction is exceeded, so object B slides on board A.

Next, find the accelerations of A and B while sliding. Let $a_A$ and $a_B$ be the rightward accelerations of A and B respectively:

\begin{aligned} &A: \quad M a_A = F - \mu' m g \\ &B: \quad m a_B = \mu' m g \end{aligned}

Computing the accelerations:

\begin{aligned} a_B &= \mu' g = 0.30 \times 9.8 = 2.94 \ \text{m/s}^2 \\ a_A &= \frac{F - \mu' m g}{M} = \frac{30 - 0.30 \times 2.0 \times 9.8}{3.0} = 8.04 \ \text{m/s}^2 \end{aligned}

The relative acceleration of B with respect to A:

a_{\text{rel}} = 8.04 - 2.94 = 5.10 \ \text{m/s}^2

The distance $L$ that B slides relative to A up to time $t=t_1$ , using the formula for uniform acceleration from rest:

L = \frac{1}{2} a_{\text{rel}} t_1^2 = \frac{1}{2} \times 5.10 \times (10)^2 = 255 \ \text{m}

Therefore, the answer is $255$ .

From the perspective of an observer on board A (non-inertial frame), the relative acceleration $\alpha$ of B can be found directly:

m \alpha = m a_A - \mu' m g

Substituting $a_A$ :

\alpha = \frac{F - \mu'(M + m)g}{M} = \frac{30 - 0.30 \times (3.0 + 2.0) \times 9.8}{3.0} = \frac{30 - 14.7}{3.0} = 5.10 \ \text{m/s}^2

Then $L = \frac{1}{2} \alpha t_1^2 = 255 \ \text{m}$ .

Friction Between Two Objects and Relative Motion