Question

A uniform metallic rod AB of length 1.4 m is lifted by two forces P and Q acting along the directions as shown in the figure. If the magnitudes of P and Q are 20 N and 50 N, respectively, find the magnitude and the position of the resultant normal force that acts on the rod.

Answer 1

The component of P along normal to AB at A = 20 `cos 60^(@)`= 10 N
The component of Q normal to AB at B = 50 cos `60^(@)` = 25 N. Thus, the given situation is similar to that shown in the figure.

Now applying the principle of moments, we get
`10(AO)=25(BO)`
` 10 (AO)=25(140-AO)`
`rArrAO=100cm` and magnitude of
` R=10N+25N=35N`