# Sign Hanging from a Cable

## Homework Statement

A 120-cm-wide sign hangs from a 5.4 kg, 200-cm-long pole. A cable of negligible mass supports the end of the rod as shown in the figure.

What is the maximum mass of the sign if the maximum tension in the cable without breaking is 380 N?

## Homework Equations

$$\Sigma \tau = 0$$

## The Attempt at a Solution

It is my understanding that the above equation is an adaptation of Newton's Second Law, $$\vec{F}=ma$$. Since the system is at equilibrium, the torque evaluates zero. I summed up the torque forces, and set them equal to zero.

$$\Sigma \tau = 0$$
$$-m_{s}gr_{s}-m_{b}(g)(r_{b})+\vec{F}(r_{c})(atan(250/200)) = 0$$

I calculated $$r_{s}$$ by taking the center of the sign and adding the initial space of 80cm from the left wall.
$$r_{s} = 0.80 + (1.20 / 2) = 1.40$$

I calculated $$r_{b}$$ by taking the length of the main bar and, since it has uniform density, simply divided the length by two.
$$r_{b} = 2 / 1 = 1$$

Plugging in the values I have, I compiled this equation:
$$-m_{s}(9.8)(1.40)-5.4(9.8)(1)+380(2)(atan(250/200)) = 0$$

Solving for $$m_{s}$$ (the mass of the sign), I get $$m_{s} \approx 46$$, but that was judged as incorrect. I believe my error may be in the $$atan(250/200)$$ at the end of my equation.

Any help would be appreciated!

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ideasrule
Homework Helper
The torque exerted by the cable is Frsin(theta). You used Fr*theta.

Alright, so I modified my equation to $$-m_{s}(9.8)(1.40)-5.4(9.8)(1)+380(2)(sin(atan(250/200))) = 0$$, which results in 18, but that is still incorrect. Is there something else I am doing wrong?

Just thought I'd add that I have made sure to use degrees... that is the correct route to go, right?

PhanthomJay