A pivot is placed at Point J and a force of 27.0N is directed downward perpendic

Question

A pivot is placed at Point J and a force of 27.0N is directed downward perpendicular to the stick at Point E. One at a time, each of the following 3 forces are applied to the stick in the manner described. Where must these forces be applied in order to achieve static equilibrium? If the is no point that will work within reason, decided which point it is (A-L) or none for each scenario.

A.) A force of 4.5N directed upward perpendicular to the stick.

B.)A force of 27.0N directed downward perpendicular to the stick.

C.) A force of 7.7N directed upward perpendicular to the stick.

Explanation / Answer

torque of the original force=27*2d =54d in clockwise direction

A) for the force of 4.5 in upward direction to prodece same torque it has to be applied at a distance of (54d/4.5)=12d to the right of j and lenghth is not suffecient to equilibrium cannot be achieved none

B)for the force of 27N in downward direction to prodece same torque it has to be applied at a distance of (54d/27)=2d to the left of j i.e. at the point D

C)for the force of 7.7 in upward direction to prodece same torque it has to be applied at a distance of (54d/7.7)=7d to the right of j i.e at the point C

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