http://www.ck12.org Chapter 8. Angular Motion and Statics
FIGURE 8.9
FIGURE 8.10
contributed by the seesaw will be zero since the center of mass of the seesaw is at the pivotP, thus simplifying our
calculations. Since the problem has unknown distance, we will use the equilibrium condition∑τ=0.
Let us first find the weights of Jessica, Boris, and the seesaw:
meg= ( 27 .kg)(10m/s^2 ) =270N
mbg= ( 33 .kg)(10m/s^2 ) =330N
msg= ( 8 .3kg)(10m/s^2 ) =83N
Jessica is 0.50 m from the left edge and therefore 2.75m− 0 .50m from the pivotP.
She contributes a (negative) clockwise torque about the pivot
PofτE=−rFsinθ= ( 2 .75m− 0 .50m)(270N)sin 90◦=−607 m·N
Boris contributes a (positive) counterclockwise torque about the pivot
PofτB=rFsinθ= (x)(330N)sin 90◦= (x)(330N)
∑τ=^0 →(−607m·N)+(330N)x=^0
x=607m330N·N= 1. 84 m
In order to balance the seesaw, Boris must sit 1.84 m from the pivot while Jessica sits 2.25 m from the pivot. Does
this seem reasonable to you, given that Boris has a greater mass than Jessica?
Check Your Understanding
What is the value ofFNin Example 1?