Answer to Question #126534 in Physics for Mwansa Kunda

Question

Three blocks of masses 10.0 kg, 5.0 kg, and 3.0 kg are connected by light strings that pass

over frictionless pulleys. The acceleration of the 5.0kg block is 2.0m/s to the left, and the surfaces are rough. Find,

i. The tension in each string.

ii. The coefficient of kinetic friction between blocks and surfaces. (Assume the same

coefficient of kinetic friction for both blocks that are in contact with surfaces).

Answer 1

Assume that tension T₁₀ acts between 10 and 5 kg blocks, tension T₃ acts between 5 and 3 kg blocks.

Applying Newton's second law, write the equation for forces acting on the 3-kg block:

$$y: -m_3g+N_3=0,\\ x: -m_3a=-T_3+\mu (m_3g).$$

For the 5-kg block:

$$y: -m_5g+N_5=0,\\ x: -m_5a=-T_{10}+T_3+\mu (m_5g).$$

For the 10-kg block:

$$y: -m_{10}g+T_{10}=-m_{10}a.$$

Thus, we have 3 equation with 3 undefined variables: $T_3, T_{10},\mu.$

$$-m_3a=-T_3+\mu (m_3g),\\ -m_5a=-T_{10}+T_3+\mu (m_5g),\\ -m_{10}g+T_{10}=-m_{10}a.$$

The solution for this system is

$$T_3=29.25\text{ N},\\ T_{10}=78\text{ N},\\ \mu=0.79.$$