Question

Box A, mass 12kg is tied to box B, mass 15kg, by a rope. Box A is pulled away from B with a force of 35N. Once moving, A experiences a force of frictions of 11N and B of 14N. What is the force of tension in the rope between the boxes?

Answer 1

Answer on Question #43263, Physics, Mechanics | Kinematics | Dynamics

Box A, mass 12kg is tied to box B, mass 15kg, by a rope. Box A is pulled away from B with a force of 35N. Once moving, A experiences a force of frictions of 11N and B of 14N. What is the force of tension in the rope between the boxes?

Solution:

After drawing the free-body diagram, we apply the equation of motion in the x-direction to get

$$\sum F _ {x} = m a$$

For box A: $m_A a = F - T - F_{frA}$

For box B: $m_B a = T - F_{frB}$

From given we have system of two equations

For box A: $12a = 35 - T - 11$

For box B: $15a = T - 14$

From second equation

$$T = 1 5 a + 1 4$$

Substitute to first equation

$$\begin{array}{l} 1 2 a = 3 5 - 1 5 a - 1 4 - 1 1 \\ 2 7 a = 1 0 \\ a = \frac {1 0}{2 7} \\ \end{array}$$

Thus,

$$T = 1 5 \cdot \frac {1 0}{2 7} + 1 4 = 1 9. 6 \mathrm {N}$$

Answer: $T = 19.6 \, \text{N}$

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