Solution to Question 18
1. Force Analysis on the Paper
Since the paper is effectively massless, the net force on it must be zero at all times ($F_{net} = m_p a \approx 0$).
The forces along the incline acting on the paper are:
- Friction from the block ($f_1$), pushing it down the incline.
- Friction from the incline ($f_2$), pushing it up the incline.
Equilibrium condition: $f_1 = f_2$.
Maximum possible friction values:
$$ f_{1,max} = \mu_1 N = \mu_1 mg \cos \theta $$
$$ f_{2,max} = \mu_2 N = \mu_2 mg \cos \theta $$
2. Case (a): $\mu_2 \ge \mu_1$
Here, the incline can provide more friction ($f_{2,max}$) than the block can exert ($f_{1,max}$).
The limiting factor is the block-paper interface.
The block slides down relative to the paper. The friction force is maxed out at $f_1 = \mu_1 mg \cos \theta$.
This force is transferred to the paper. The paper requires $f_2 = f_1 = \mu_1 mg \cos \theta$ to stay in equilibrium.
Since $\mu_2 \ge \mu_1$, the incline can provide this force ($f_2 < f_{2,max}$). Therefore, the paper does not slip relative to the incline.
Accelerations:
- Paper: $a_p = 0$.
- Block: $mg \sin \theta – f_1 = m a_b \implies a_b = g(\sin \theta – \mu_1 \cos \theta)$.
Since the paper is effectively massless, the net force on it must be zero at all times ($F_{net} = m_p a \approx 0$).
The forces along the incline acting on the paper are:
- Friction from the block ($f_1$), pushing it down the incline.
- Friction from the incline ($f_2$), pushing it up the incline.
Maximum possible friction values: $$ f_{1,max} = \mu_1 N = \mu_1 mg \cos \theta $$ $$ f_{2,max} = \mu_2 N = \mu_2 mg \cos \theta $$
Here, the incline can provide more friction ($f_{2,max}$) than the block can exert ($f_{1,max}$).
The limiting factor is the block-paper interface.
The block slides down relative to the paper. The friction force is maxed out at $f_1 = \mu_1 mg \cos \theta$.
This force is transferred to the paper. The paper requires $f_2 = f_1 = \mu_1 mg \cos \theta$ to stay in equilibrium.
Since $\mu_2 \ge \mu_1$, the incline can provide this force ($f_2 < f_{2,max}$). Therefore, the paper does not slip relative to the incline.
Accelerations:
- Paper: $a_p = 0$.
- Block: $mg \sin \theta – f_1 = m a_b \implies a_b = g(\sin \theta – \mu_1 \cos \theta)$.
Here, the block can exert more friction than the incline can withstand. The “weak link” is the paper-incline interface.
The paper will slip on the incline. The resisting friction is maxed out at $f_2 = \mu_2 mg \cos \theta$.
Since $f_1 = f_2$ (massless paper), the friction acting on the block is also limited to $\mu_2 mg \cos \theta$, even though $\mu_1$ allows for more. This means the block and paper do not slip relative to each other; they slide together down the incline.
Accelerations:
Both move with common acceleration $a$.
$$ mg \sin \theta – f_2 = m a $$
$$ a = g(\sin \theta – \mu_2 \cos \theta) $$
Thus, $a_b = a_p = g(\sin \theta – \mu_2 \cos \theta)$.
(a) $a_b = g(\sin \theta – \mu_1 \cos \theta), \quad a_p = 0$
(b) $a_b = a_p = g(\sin \theta – \mu_2 \cos \theta)$