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.

Sketch of the problem

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. The part where I'm strugglin the most is where are the forces acting. Therefore I need help with the free body diagram of the forces here. Can someone help me here please?.


1 Answer 1


Let the sphere exert a force $F$ on the rod perpendicular to it. We know that the vertical component of this force will be we equal to the weight of the sphere. $$F\times\cos37^o=10\times 9.8$$ Now, let the component of the weight of the rod in perpendicular direction be equal to $F'$. $$F'=5\times 9.8\times\cos37^o$$ We know that the system is in equilibrium. This implies that $\tau_{net}=0$ on the rod. This gives us our final equation which gives the value of tension $T$. $$\frac {F+F'}2=T$$


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