Problem: A uniform rod of mass M = 13.0 kg and length L = 1.80 m is held horizontally with one end anchored in the wall by a hinge that allows the rod to rotate freely in the x-y plane. The other end of the rod is attached to the wall with cable that makes angle θ = 80◦ with the rod. Magnitude of acceleration due to gravity is g = 10 m/s2. What is the magnitude of tension force FT in the cable?

The magnitude of the tension force is  

Explanation:

The diagram of the question is shown on the first uploaded image

     From the question we are told that

         The mass of the rod is

          The length of the rod is

          The angle the cable (string) makes with the rod is  

          The acceleration due to gravity is  

Torque is mathematically represented as

There are two torques acting on the rod

            The first is torque due to gravitational force

 This force is mathematically represented as

The  distance is

The angle is  

   The first  torque due to gravitational force

The second  is torque due to the tension on the cable

 The tension on the string is

 Since this force is  acting at the other end of the rod , the is  from that point to the point of torque(rotation ) is  

    The angle which   made with the  point it is acting on is given as

 The second  torque due to tension in the string is

At equilibrium

Making   the subject of the formula  we have  

Substituting values

you have to draw the persons position over there

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