Question

A piece of modern sculpture consists of an 8.0-m-long, 150 kg stainless steel bar passing diametrically through a 50 kg copper sphere. The center of the sphere is 2.0 m from one end of the bar. To be mounted for display, the bar is oriented vertically, with the copper sphere at the lower end, then tilted 35° from vertical and held in place by one horizontal steel cable attached to the bar 2.0 m from the top end. What is the tension in the cable?

Short answer

Tension in the cable is T= 800.6 N

Step-by-step solution

Given information

Length of bar = 8.0 m
mass of bar =150 kg
The bar is attached with a 50 kg at 2 m from the end of the bar.
The rod is tilted by 35^o from vertical and attached with a cable 2 m from top end.

Explanation

We will equate the torque and then solve to get the tension in the cable

First find the perpendicular distance in order to calculate torque

Distance of force${\left({\overrightarrow{F}}_G\right)}_{S\text{phere}}\mathrm{is}2mSin35°$

Distance of force ${\left({\overrightarrow{F}}_G\right)}_{\text{bar}}\mathrm{is}4mSin35°$

Distance of force $\left(\overrightarrow{T}\right)is\left(6m\right)Cos35°$
Solve for T

$$\begin{gathered} =\frac{50\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2}{6\mathrm{m}}2m\tan 35°+\frac{150\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2}{6\mathrm{m}}4m\tan 35° \\ \mathrm{T}=800.6\mathrm{N} \end{gathered}$$