Sum no. 14

Sum no. 14 A weight 10 gf at

Dear Student,

Please find below the solution to the asked query:

Given that the force on the left side of O is F = 150 g f. This force is acting at a distance of 40 cm from O.

Therefore, the Anti clock wise torque is,

${\tau }_{ACW}=Fr \sin \theta =150\times 10\times 0.4\times \sin 90=600 N-m$

Now the force on the right hand side to O is 250 gf and this force is acting at a distance of 20 cm. Therefore, the clockwise direction is,

${\tau }_{CW}=Fr \sin \theta =250\times 10\times 0.2\times \sin 90=500 N-m$

Therefore, the difference of the torques is,

${\tau }_{diff}={\tau }_{ACW}-{\tau }_{CW}=600-500=100 N-m$

If the rod has to be balanced, then the extra force of 100 gf​ has to apply at a distance x on the right hand side to O. Therefore,

${\tau }_{diff}=100\times 10\times x\times \sin 90\Rightarrow 100=1000x\Rightarrow x=10 cm$

Hope this information will clear your doubts about the topic.

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