Question

Three objects are connected on a table as shown in above figure. The coefficient of kinetic friction between the block of mass $$m_{2}$$ and the table is $$0.350$$. The objects have masses of $$m_{1} = 4.00 kg, m_{2} = 1.00 kg$$, and $$m_{3} =2.00kg$$ and the pulleys are frictionless. (a) Draw a free body diagram of each object. (b) Determine the acceleration of each object, including its direction. (c) Determine the tensions in the two cords. What If? (d) If the tabletop were smooth, would the tensions increase, decrease, or remain the same? Explain.

Answer 1

(a) The free-body diagrams for each object are given below.

(b) Let $$a$$ represent the positive magnitude of the acceleration $$−a\hat{j}$$ of $$m_{1}$$, of the acceleration $$−a\hat{i}$$ of $$m_{2}$$, and of the acceleration $$+a\hat{j}$$ of $$m_{3}$$. Call $$T_{12}$$ the tension in the left cord and $$T_{23}$$ the tension in the cord on the right.
For $$m_{1}, \sum F_{y} = ma_{y} : +T_{12} − m_{1}g = −m_{1}a$$
For $$m_{2}, \sum F_{x} = ma_{x} : −T_{12} + \mu_{k}n + T_{23} = −m_{2}a$$
and $$\sum F_{y} = ma_{y}$$, giving $$n −m_{2}g = 0$$.
For $$m_{3}, \sum F_{y} = ma_{y}$$, giving $$T_{23} −m_{3}g = +m_{3}a$$.
We have three simultaneous equations:
$$−T_{12} + 39.2 N = (4.00 kg)a$$
$$+T_{12} − 0.350(9.80 N) − T_{23} = (1.00 kg)a$$
$$+T_{23} − 19.6 N = (2.00 kg)a$$
Add them up (this cancels out the tensions):
$$+39.2 N − 3.43 N − 19.6 N = (7.00 kg)a$$
$$a = 2.31 m/s^{2}$$, down for $$m_{1}$$, left for $$m_{2}$$, and up for $$m_{3}$$

(c) Now $$−T_{12} + 39.2 N = (4.00 kg)(2.31 m/s^{2} )$$
$$T_{12} = 30.0 N$$
and
$$T_{23} − 19.6 N = (2.00 kg)(2.31 m/s^{2} )$$
$$T_{23} = 24.2 N$$

(d) If the tabletop were smooth, friction disappears $$(\mu_{k} = 0)$$, and so the acceleration would become larger. For a larger acceleration, according to the equations above, the tensions change:
$$T_{12} = m_{1}g − m_{1}a\rightarrow T_{12}$$ decreases.
$$T_{23} = m_{3}g + m_{3}a\rightarrow T_{23}$$ increases.