Question

1.      A system comprising blocks, a light frictionless pulley, a frictionless incline, and connecting ropes is shown in the figure. The 9.0-kg block accelerates downward when the system is released from rest. What is the tension in the rope connecting the 6.0-kg block and the 4.0-kg block? [42 N]

6.0 kg 9.0 kg 30

Answer 1

The way to solve these problems is to identify all the forces on each block and apply newtons second law ( m a = Sum F )

The block of 6 kg:
+ two forces in the direction of motion:
1. gravity along the slope - m g sin(30)
2. Tension in the rope between the masses on the slope: T1
So net force in the direction of motion: T1 - m g sin(30)
=> 6 a = T1 - 6 * g * 1/2
=> 6 a = T1 - 3 g

The block of 4 kg:
+ three forces in the directin of motion:
1. gravity along slope - 4 g sin(30)
2. tension of the rope between the 6 kg block and 4 kg block -T1
3 tension of the rope between the 4kg block and the 9 kg block T2

So the net force in the direction of motion is T2 - T1 - 4 g sin(30)
=> 4a = T2 - T1 - 2g

The block of 9 kg:
+ two forces in the direction of its motion:
1. Tension opposing its motion: -T2
2. force of gravity: m g
So the net force in the direction of motion is 9 g - T2
=> 9 a = 9 g - T2


Now note that we have obtained three equations for the three unknowns a, T1 and T2:

6a = T1 - 3g

4a = T2 - T1 - 2g

9a = 9g - T2

In principle we can solve this set of equations for all the unknowns, but for this problem we only need to solve for T1:

The third equations gives

T2 = 9 g - 9 a

substituting that in the second equation gives

4 a = 9 g - 9 a - T1 - 2 g
i.e.
13 a = 7 g - T1

So we have reduced the problem to two equations in two unknowns:

6a = T1 - 3g
13 a = 7 g - T1

or, upon multiplying the first equation with 13 and the second with 6:

78 a = 13 T1 - 39 g
78 a = 42 g - 6 T1

subtracting the second from the first eliminates a:

0 = 19 T1 - 81 g

T1 = 81/19 * g

With g = 9.8 m/s^2 this gives

T1 = 41.8 N = 42 N