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

Jun 2017
399
6
Lima, Peru
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?.
 

skeeter

Math Team
Jul 2011
3,356
1,848
Texas
$\color{red}Forces$ acting on the bar ... $\color{blue}Forces$ acting on the sphere

I get T = 71.25 N

Sphere_bar.jpg
 
Jun 2017
399
6
Lima, Peru
$\color{red}Forces$ 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
Jul 2011
3,356
1,848
Texas
Sphere_eq.jpg
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 ...

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

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