Question

$M_A=3kg,M_B=4kg,$ and $M_C=8kg.\mu$ between any two surfaces is 0.25. Pulley is frictionless and string is massless. A is connected to wall through a massless rigid rod.

• A. the value of F to keep C moving with constant speed is 80 N
• B. the value of F to keep C moving with constant speed is 120 N
• C. if F is 200 N then acceleration of B is 10 ms${}^{-2}$
• D. to slide C towards left, F should be at least 50 N (Take g = 10 ms${}^{-2}$)

Answer 1

Answer: (A), (C)

the value of F to keep C moving with constant speed is 80 N
if F is 200 N then acceleration of B is 10 ms${}^{-2}$

Lest calculate frictional forces first.

Frictional force between A and B

$f_{AB}=\mu M_{A}g$

$⟹f_{AB}=\frac{30}{4}$ N

Frictional force between B and C

$f_{BC}=\mu (M_A+M_B)g$

$⟹f_{BC}=\frac{70}{4}$ N

Friction force between C and ground

$f_{CG}=\mu (M_A+M_B+M_C)g$

$⟹f_{CG}=\frac{150}{4}$ N

If C is moving then tension in string will be equal to total frictional forces on B

$T=f_B=f_{AB}+f_{BC}$

$⟹T=f_B=\frac{100}{4}$ N. . . . . . .(1)

If C is moving then frictional force on it

$f_C=f_{CG}+f_{BC}$

$⟹f_C=\frac{220}{4}$ N

External force F should be equal to frictional force on C plus tension in the string.

$⟹F=\frac{220}{4}+\frac{100}{4}$

$⟹F=80$ N

If $F=200N$

Total frictional forces on B and C ( calculated above)

$f=120N$

Force that moving block B and C:

$F^{\prime }=appliedforce-frictionalforces$

$⟹F^{\prime }=$ 200 - 80 = 120 N

Acceleration of B and C

$a=\frac{F^{\prime }}{M_C+M_B}$

$⟹a=\frac{120}{8+4}$

$⟹a=10m{s}^{-1}$