A horizontal plank 4.00m long and having mass 20.0kg rests on two supports, one at the far left end, and a second 1.00m from the right end. Find the magnitude of the force exerted on the plank by the second pivot. (Hint: If you use the far left end as the point from which you will calculate the torques about it will simplify the problem.) Remember equilibrium conditions mean that sum of torques and forces equals zero.

The magnitude of the force exerted on the plank by the second pivot is; 131 N

We are given;

Mass of plank; m = 20 kg

Length of plank; L = 4 m

Distance of second pivot from left end of plank = 3 m

Now, I have attached a picture of this plank with F representing the force from the weight of the plank acting downwards.

Now, taking moments about the left end which is the first pivot we have;

(F2 × 3) - (F × 2) = 0

Where;

F2 is the force exerted on the plank by the second pivot

F is Force by the weight of the plank acting at the center downwards;

F = mg

F = 20 × 9.8

F = 196 N

Thus;

(F2 × 3) - (196 × 2) = 0

3F2 - 392 = 0

F2 = 392/3

F2 = 130.67

F2 ≈ 131 N

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Force in the second pivot = 130.8 [N]

Explanation:

This problem can be solved by means of a static analysis, performing an equation of sum of moments around a point equal to zero and clearing the unknown in that equation we can reach the solution of the problem.

But first we must make a sketch of the problem as well as a free body diagram.

In the attached diagram we can see the free body diagram with the respective forces and with the developed equation.

The mass and therefore the weight of the plank are concentrated in half of this, we make a sum of moments in Pivot A so that the equation of moments is a function of RB, which corresponds to the second pivot. The equation can be seen in the attached diagram.

i think the answer is a, hope this : )