A uniform rod has a weight of 40 N and a length of 1 m. It is hinged to a wall [at the left end]and held in a horizontal position by a vertical massless string[at the right end]. Find:
1.Force in the string 
2.Magnitude of hinge force

Dear student ,
Here in this case the sum of torque is equal to zero .
So , τ=-r1×F1 + r2×F2=0r1×F1=r2×F2τ2=0·5×40×sin90°=20 Nm
Now look this is the magnitude of torque exerted by the string about a horizontal axis and perpendicular to the torque .
Now the force on the string is the is ,
F=torquelength of the rod=τ2l=20 Nm1 m=20 N
Now to find the three forces look there are only three forces acting on the rod and this three forces are vertical .
Now the hinge force is , F'=Torquelength of the rod=201=20 N

Regards

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