Force exerted on a pivot point

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    Force Pivot Point
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To determine the force exerted on a 4.0 m horizontal plank by the second pivot point, the problem involves analyzing the plank's equilibrium. The total weight of the plank, which is 24.0 kg, must be balanced by the forces at the two pivot points. By taking moments around a convenient point, it can be established that the torques from each pivot are equal and opposite. The center of mass is assumed to be at the midpoint of the plank, 2.0 m from either end. This analysis leads to the conclusion that the forces at the pivots must sum to equal the weight of the plank, ensuring both translational and rotational equilibrium.
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Homework Statement


A horizontal plank 4.0 m long and having mass 24.0 kg rests on two pivot points, one left at the end and a second one 1.0 m from the right end. What is the magnitude of the force exerted on the plank by the second pivot point?

Homework Equations

The Attempt at a Solution


I'm not even sure how to attempt this problem
 
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Draw a large diagram and mark on the forces and their location.

Take moments about a convenient point.
 
Draw a free body diagram.
Assuming the mass distribution is same, take the centre of mass at the centre i.e at 2.0 metre.
The plank is in translational and rotational equilbrium.
Thus, the sum of magnitude of the forces at the pivot is equal to the wieght of the plank.
And, the moments, or the torque exerted by each of the pivot is equal and opposite thus their magnitudes are equal.
Take the origin as the centre of mass.

PS:- Torque= Force*distance of action from the origin (when force is perpendicular)
In vector notation torque is the cross product of radius and force. But the above equation is enough here

Hope it helps :)
 
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