Analysis of Forces

The force $F$ acts on the pulley, so the tension in the string is $T = F/2$.

• Block A: Pulled right by $F/2$. Friction $f_1$ acts Left (opposing motion relative to B).
• Block C: Pulled right by $F/2$. Friction $f_2$ acts Left (opposing motion relative to B).
• Block B: Dragged Right by friction from A ($f_1$) and friction from C ($f_2$).

Case 1: All blocks move together (No Slipping)

Common acceleration $a$. System mass $= 3m$. Net force $= F$.

Conditions for Static Friction:

For Block A: $F_{net} = T – f_1 = ma$. $$ \frac{F}{2} – f_1 = m\left(\frac{F}{3m}\right) = \frac{F}{3} \implies f_1 = \frac{F}{2} – \frac{F}{3} = \frac{F}{6} $$ Required condition: $f_1 \le \mu mg \implies \frac{F}{6} \le \mu mg \implies \mathbf{F \le 6\mu mg}$.

(Note: Check Block C. $f_2 = F/6$ as well. Limit is $2\mu mg$. Block A slips first.)

Result: $a_B = \frac{F}{3m}$

Case 2: A slips, but B and C move together

Range: $F > 6\mu mg$. Friction at A is kinetic: $f_1 = \mu mg$.

Block A accelerates at $a_A = (F/2 – \mu mg)/m$. Friction $f_1 = \mu mg$ acts to the right on Block B.

Now consider the system (B + C). Total mass $2m$. Forces:

• Driving Force on C: $T = F/2$ (Right)
• Driving Force on B (from A): $f_1 = \mu mg$ (Right)

Check Validity (No slip between B and C):

Force eq for C: $T – f_2 = m a_{BC}$. $$ \frac{F}{2} – f_2 = m \left( \frac{F + 2\mu mg}{4m} \right) \implies f_2 = \frac{F}{4} – \frac{\mu mg}{2} $$ Condition: $f_2 \le 2\mu mg$. $$ \frac{F}{4} – \frac{\mu mg}{2} \le 2\mu mg \implies \frac{F}{4} \le 2.5 \mu mg \implies \mathbf{F \le 10\mu mg} $$

Result: $a_B = \frac{F + 2\mu mg}{4m}$

Case 3: All interfaces slip

Range: $F > 10\mu mg$. Friction is kinetic everywhere.

• Force on B from A ($f_1$): $\mu mg$ (Right)
• Force on B from C ($f_2$): $2\mu mg$ (Right)

Checking accelerations: $a_C \approx F/2m$ vs $a_B \approx 3\mu g$. For large F, $a_C > a_B$. So C slides forward relative to B. Friction on B is forward.