Question

A 120-cm-wide sign hangs from a 5.0 kg, 200-cm-long pole. A cable of negligible mass supports the end of the rod as shown in Figure P12.62. What is the maximum mass of the sign if the maximum tension in the cable without breaking is 300 N?

Short answer

Maximum mass it can support is 30.54 kg

Step-by-step solution

Given Information

A 120-cm-wide sign hangs from a pole.

Mass of pole = 5.0 kg,

Length of pole 200-cm= 2m

A cable supports the end of the rod with maximum tension= 300 N

Explanation

First draw the diagram to understand and solve the problem as below

Now equate the clockwise and anti-clockwise torque

$mg\times \left(80cm\right)+\left(5kg\right)g\times \left(100cm\right)+mg\times \left(200cm\right)=TSin\theta \times \left(200cm\right)$

Substitute

$sin\theta =\frac{250}{\sqrt{({250}^2+{200}^2)}}=0.78and\mathrm{T}=300\mathrm{N}$

We get,

$$\begin{gathered} mg\times (0.80m)+\left(5kg\right)g\times \left(1m\right)+mg\times \left(2m\right)=\left(300N\right)(0.78)\times \left(2m\right) \\ 2.8m+5kg=\frac{468}{g}kg \\ m=15.27kg \end{gathered}$$

Maximum mass of sign board = M =2m =30.54 kg.