Question

How do I solve this problem Determine the weight of the tram. The cable at left exerts a 30,000 N force. (The tram is attached to the cable so the tension in the left cable is not necessarily equal to the tension in the right cable.)

Answer 1

Free body diagram:

Here, T_1 is the tension in the left cable (T_1=30000 N; given), T_2 is the tension in the right cable, W is the weight of the tram.

The force equation in horizontal direction is given as,

`T_1=T_2\sin (75\degree)`

Therefore, the tension in right cable T_2 is given as,

`T_2=(T_1)/(\sin(75\degree))`

Substituting all known values,

`\begin{gathered} T_2=\frac{30000\text{ N}}{\sin (75\degree)} \\ \approx31058.28\text{ N} \end{gathered}`

The force equation in the vertical direction is given as,

`W=T_2\cos (75\degree)`

Substituting all known values,

`\begin{gathered} W=(31058.28\text{ N})*\cos (75\degree) \\ \approx8038.47\text{ N} \end{gathered}`

Therefore, the weight of the tram is 8038.47 N.