# How to find the modulus of the tension of a wire when a sphere is over a bar?

#### Chemist116

The problem is as follows:

The figure from below shows a system which is at equilibrium. The sphere is frictionless, homogeneous and its mass is $5\,kg$. The homogeneous bar has a mass of $10\,kg$. Find the modulus of the tension of the wire. Assume that the acceleration due gravity is $10$ meters per second square. M is the midpoint of the bar. The alternatives are as follows:

$\begin{array}{ll} 1.&62N\\ 2.&71.25N\\ 3.&80.05N\\ 4.&92.25N\\ 5.&100N\\ \end{array}$

Can someone help me here?. I'm not exactly how to apply the condition of equlibrium for this problem. I'm confused exactly how should I use the information of the angle given at one end of the wire.

What I've attempted to do was the following:

$-Tl+100\sin 53^{\circ}\left(\frac{l}{2}\right)+50\sin 53^{\circ}\left(\frac{l}{2}\right)=0$

But this did not solved the problem. CThe part where I'm strugglin the most is where are the forces acting. Can someone help me here please?.

#### Chemist116

$\color{red}Forces$ acting on the bar ... $\color{blue}Forces$ acting on the sphere

I get T = 71.25 N

View attachment 10937
I see a bunch of forces but how do I use them in an equation?. Do you refer a torque about the pivot?.

$-T(l)+Mg\left(\frac{l}{2}\sin 53^{\circ}\right)+mg\frac{l}{2}\left(\frac{4}{5}\right)=0$

This would become into:

$-T+\frac{100}{2}\left(\frac{4}{5}\right)+50\frac{1}{2}\left(\frac{4}{5}\right)=0$

$T=60\,N$

What went wrong here?.

#### skeeter

Math Team Starting with the equilibrium condition on the sphere, note $\color{blue}mg = R_{bs} \cos{\theta} \implies R_{bs} = \dfrac{mg}{\cos{\theta}} = \dfrac{25g}{4}$.

Also note the reaction force of the bar on the sphere at point M equals the reaction force of the sphere on the bar, i.e. ${\color{blue}R_{bs}}={\color{red}R_{sb}}$

Using the bar's pivot on the wall, the rotational equilibrium of the bar can be summarized with the equation ...

$\color{red}T \cdot L = R_{sb} \cdot \dfrac{L}{2} + Mg \cos{\theta} \cdot \dfrac{L}{2}$

$T = \dfrac{25g}{8} + 40 = 71.25 \, N$