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Question

Two blocks connected by a rope of negligible mass are being dragged by a horizontal force (Fig. above). Suppose $$F = 68.0 N, m_{1} = 12.0 kg, m_{2} = 18.0 kg$$, and the coefficient of kinetic friction between each block and the surface is $$0.100$$. (a) Draw a free-body diagram for each block. Determine (b) the acceleration of the system and (c) the tension $$T$$ in the rope.

Solution
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Because the cord has constant length, both blocks move the same number of centimeters in each second and so move with the same acceleration. To find just this acceleration, we could model the $$30kg$$ system as a particle under a net force. That method would not help to finding the tension,
so we treat the two blocks as separate accelerating particles.

(a) Below figure shows the free-body diagrams for the two blocks.
The tension force exerted by block $$1$$ on block $$2$$ is the same size as the tension force exerted by object $$2$$ on object $$1$$. The tension in a light string is a constant along its length, and tells how strongly the string pulls on objects at both ends.

(b) We use the free-body diagrams to apply Newton’s second law.
For $$m_{1}: \sum F_{x} = T − f_{1} = m_{1}a$$ or $$T = m_{1}a + f_{1}$$........... [1]
And also
$$\sum F_{y} = n_{1} − m_{1} g = 0$$ or $$n_{1} = m_{1}g$$
Also, the definition of the coefficient of friction gives
$$f_{1} = \mu n_{1} = (0.100)(12.0 kg)(9.80 m/s^{2}) = 11.8 N$$
For $$m_{2}: \sum F_{x} = F – T – f_{2} = ma$$...................................... [2]
Also from the $$y$$ component, $$n_{2} – m_{2}g = 0$$ or $$n_{2} = m_{2}g$$
And again $$f_{2} = \mu n_{2} = (0.100)(18.0 kg)(9.80 m/s^{2}) = 17.6 N$$
Substituting $$T$$ from equation [1] into [2], we get
$$F − m_{1}a − f_{1} − f_{2} = m_{2}a$$ or $$F − f_{1} − f_{2} = m_{2}a + m_{1}a$$
Solvind for $$a$$
$$a=\frac{F-f_{1}-f_{2}}{m_{1}+m_{2}}=\frac{(68.0N-11.8N-17.6N)}{(12.0kg+18.0kg)}=1.29m/s^{2}$$

(c) From equation [1]
$$T=m_{1}a+f_{1}=(12.0kg)(1.29m/s^{2})+11.8N=27.2N$$

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