A woman of mass m = 50.9 kg sits on the left end of a seesawa plank of length L

Question

A woman of mass m = 50.9 kg sits on the left end of a seesawa plank of length L = 3.99 m, pivoted in the middle as shown in the figure.

A woman of mass m = 50.9 kg sits on the left end of a seesawa plank of length L = 3.99 m, pivoted in the middle as shown in the figure. (a) First compute the torques on the seesaw about an axis that passes through the pivot point. Where should a man of mass M = 71.2 kg sit if the system (seesaw plus man and woman) is to be balanced? m (b) Find the normal force exerted by the pivot if the plank has a mass of mpl = 13.4 kg. N (c) Repeat part (a), but this time compute the torques about an axis through the left end of the plank. The system consists of two people and a seesaw. Because the sum of the forces and the sum of the torques acting on the system are both zero, the system is said to be in equilibrium. Goal Apply the conditions of equilibrium

Explanation / Answer

Part A)

About the center pivot point, the torque for the woman is found by...

= FL = (mg)L = (50.9)(9.8)(2) = 997.64 Nm

The torque from the man must equal that to balance the seesaw.

997.64 = (mg)L = (71.2)(9.8)(L)

L = 1.43 m to the right of the pivot point according to the picture.

The torque from the mass of the see-saw and from the fulcrum cancels out (equals zero) since = FL and L is zero (L is the distance to the pivot point from the weight of the see saw / fulcrum, the weight is concentrated at the center and the fulcrum is at the center also, so that location is the pivot point, so L = 0)

Part B)

The pivot point supports the weight of all three items (man, woman, and see-saw)

Fn = (m + m + m)g

Fn = (50.9 + 71.2 + 13.4)(9.8)

Fn = 1327.9 N

Part C)

The torque from the woman = 0 since there is zero distance from her to the pivot point since she is at the pivot point

The torque from the man = FL = (mg)L = (71.2)(9.8)(2 + 1.43) = 2393.3 Nm

The torque from the weight of the see-saw = (mg)L = (13.4)(9.8)(2) = 262.64 Nm

The torque from the fulcrum = (Fn)L = (1327.9)(2) = -2655.8 Nm (The negative is sign convention since it is counterclockwise. You can ignore it if needed for your answer key)

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