Solution to Question 38

(a) Condition for Block B not sliding relative to the paper

This implies the Paper and Block B move together. For the paper not to slide under B, the friction required to accelerate B must be static.

1. Maximum Accelerations:

• Max acceleration of A (driven by friction from paper): $$ f_{A,max} = \mu mg \implies a_{A,max} = \frac{\mu mg}{m} = \mu g $$
• Max acceleration of B (driven by friction from paper): $$ f_{B,max} = \mu N_{paper-B} \approx \mu mg $$ (Note: $N_{paper-B} \approx mg$ because paper is massless and only supports A). $$ a_{B,max} = \frac{f_{B,max}}{M} = \frac{\mu mg}{M} = \mu g \left(\frac{m}{M}\right) $$

Since $m < M$, $a_{B,max} < a_{A,max}$. Block B is the bottleneck; it will slip first.

2. Limiting Force:

For B not to slide, the system acceleration $a$ must not exceed $a_{B,max}$. $$ a \le \mu g \frac{m}{M} $$ The force required to accelerate the whole system (A + Paper + B) at this rate is: $$ F = (M + m) a \le (M+m) \mu g \frac{m}{M} $$ $$ F \le \mu mg \left( 1 + \frac{m}{M} \right) $$

(b) Paper slides relative to B, but not A

This implies:

1. Paper slips on B: Friction on B is limiting kinetic ($f_B = \mu mg$). B accelerates at $a_B = \mu g \frac{m}{M}$.
2. Paper does not slip on A: A moves with the paper. $a_A = a_{paper} = a$.

Equation of motion for Paper (massless): $$ F = f_A + f_B $$ Since A is moving with acceleration $a$: $f_A = ma$. Since B is slipping: $f_B = \mu mg$. $$ F = ma + \mu mg $$

Condition for A not slipping: The required friction $f_A$ must be less than $f_{A,max}$. $$ ma \le \mu mg \implies a \le \mu g $$ Substitute max $a$ into the Force equation: $$ F_{max} = m(\mu g) + \mu mg = 2\mu mg $$

So the range starts from the limit of case (a) and ends at $2\mu mg$.