Question

In the figure shown below, the mass of block A, B, and C are 7kg, 4kg and 3 kg respectively. Coefficient of friction between A and B is 0.4, coefficient of friction between B and C is 0.3. There is no friction between ground and A. Force F of 70N is applied on block B rightward. Find the direction of friction force on block A and C respectively.

• A. Leftward, rightward.
• B. Rightward, rightward
• C. Leftward, leftward
• D. Rightward, leftward

Answer 1

Answer: (B)

Rightward, rightward

Answer 2

Let's analyze the forces and potential motion of the blocks. The masses are $m_A = 7 \, \text{kg}$, $m_B = 4 \, \text{kg}$, $m_C = 3 \, \text{kg}$. The coefficient of friction between A and B is $\mu_1 = 0.4$. The coefficient of friction between B and C is $\mu_2 = 0.3$. The applied force on B is $F = 70 \, \text{N}$ rightward. There is no friction between the ground and A. Let $g = 10 \, \text{m/s}^2$.

First, consider the interaction between B and C. The normal force between B and C is $N_{BC} = m_C g = 3 \times 10 = 30 \, \text{N}$. The maximum static friction between B and C is $f_{s,BC}^{max} = \mu_2 N_{BC} = 0.3 \times 30 = 9 \, \text{N}$. If C moves with B, the friction force required on C is $f_C$. Assuming they move together with acceleration $a$, $f_C = m_C a$.

Consider the interaction between A and B. The normal force between A and B is $N_{AB} = (m_B + m_C) g = (4 + 3) \times 10 = 70 \, \text{N}$. The maximum static friction between A and B is $f_{s,AB}^{max} = \mu_1 N_{AB} = 0.4 \times 70 = 28 \, \text{N}$. If A moves with B, the friction force required on A is $f_A$. Assuming they move together with acceleration $a$, $f_A = m_A a$.

Let's assume all three blocks move together with a common acceleration $a$. The total mass of the system is $M = m_A + m_B + m_C = 7 + 4 + 3 = 14 \, \text{kg}$. The net horizontal force on the system is $F = 70 \, \text{N}$. So, $F = M a \implies 70 = 14 a \implies a = 5 \, \text{m/s}^2$.

For block C to move with this acceleration, the friction force on C from B must be $f_{BC} = m_C a = 3 \times 5 = 15 \, \text{N}$. The maximum static friction between B and C is $f_{s,BC}^{max} = 9 \, \text{N}$. Since the required friction force (15 N) is greater than the maximum static friction (9 N), block C will slip relative to block B. The friction between B and C is kinetic friction, and its magnitude is $f_{k,BC} = f_{s,BC}^{max} = 9 \, \text{N}$. Block B is pushed rightward. It tends to move right relative to C. So, the friction force on C from B is in the direction of motion of B relative to C, which is rightward (to drag C along). Thus, the friction force on C is rightward.

Now consider the motion of A and B. Since C slips on B, the acceleration of B is not necessarily 5 m/s$^2$. Let's consider the system A+B. The forces acting horizontally on the system A+B are the applied force F on B, and the friction force from C on B. The friction force from C on B is opposite to the friction force from B on C, so it is $f_{BC} = 9 \, \text{N}$ to the left. The net horizontal force on the system A+B is $F - f_{BC} = 70 - 9 = 61 \, \text{N}$. If A and B move together with acceleration $a_{AB}$, then $F - f_{BC} = (m_A + m_B) a_{AB} \implies 61 = (7 + 4) a_{AB} \implies 61 = 11 a_{AB} \implies a_{AB} = 61/11 \, \text{m/s}^2$. For block A to move with this acceleration, the friction force on A from B must be $f_{AB} = m_A a_{AB} = 7 \times (61/11) = 427/11 \approx 38.82 \, \text{N}$. The maximum static friction between A and B is $f_{s,AB}^{max} = 28 \, \text{N}$. Since the required friction force ($\approx 38.82$ N) is greater than the maximum static friction (28 N), block A will slip relative to block B. The friction between A and B is kinetic friction, and its magnitude is $f_{k,AB} = f_{s,AB}^{max} = 28 \, \text{N}$. Block B is pushed rightward and tends to move right relative to A. So, the friction force on A from B is in the direction of motion of B relative to A, which is rightward (to drag A along). Thus, the friction force on A is rightward.

The direction of friction force on block A is rightward. The direction of friction force on block C is rightward.

Explanation of the solution:

1. Assume all blocks move together with the same acceleration. Calculate the required friction forces for this to happen.
2. Compare the required friction forces with the maximum possible static friction forces at each interface.
3. If the required friction exceeds the maximum static friction, slipping occurs, and the friction is kinetic, with magnitude equal to the maximum static friction. Determine the direction of kinetic friction based on the direction of relative motion.
4. In this case, both C slips on B, and A slips on B.
5. Friction on C from B is kinetic. B tends to move right relative to C, so friction on C is rightward.
6. Friction on A from B is kinetic. B tends to move right relative to A, so friction on A is rightward.