Let\'s begin by determining the equilibrium position of a seesaw pivot. You and
ID: 1466457 • Letter: L
Question
Let's begin by determining the equilibrium position of a seesaw pivot. You and a friend play on a seesaw. Your mass is 90 kg, and your friend’s mass is 60 kg. The seesaw board is 3.0 m long and has negligible mass. Where should the pivot be placed so that the seesaw will balance when you sit on the left end and your friend sits on the right end?
SOLUTION
SET UP First we sketch the physical situation (Figure 1) , and then we make a free-body diagram (Figure 2) , placing the pivot at a distance x from the left end. Note that each weight is the corresponding mass multiplied by g, so g will divide out of the final result. Remember that counterclockwise torques are positive and clockwise torques are negative. For the seesaw to be in equilibrium, the sum of the torques about any axis must be zero. We choose to take torques about the pivot axis; then the normal force acting at the pivot doesn't enter into the analysis. The moment arms are x for you and (3.0m–x) for your friend.
SOLVE Your torque is ?you=+(90kg)gx. Your friend is 3.0m?x from the pivot, and the corresponding torque is ?friend=?(60kg)×g(3.0m?x). For equilibrium, the sum of these torques must be zero:
?you+?friend+(90kg)gx?(60kg)g(3.0m?x)==00
We divide through by g and solve for x, obtaining x=1.2m.
REFLECT If we put the pivot at the center of the board, where x=1.5m, we would have equal moment arms but unequal weights, so it's clear that the pivot must be to the left of center.
Part A - Practice Problem:
If the board has a mass of 16 kg, where should the pivot be placed for balance?
Express your answer in meters to two significant figures.
60 kg go kg 3.0 mExplanation / Answer
For equilibrium, the sum of all the torques must be zero
90*x=60*(3-x)+16*(1.5-x)
x= 1.22 m
Pivot must be placed at 1.22 m from 90 kg end
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