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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» Solution :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` ![]() |
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