The seesaw in the figure below is 4.5 m long. Its mass of 20 kg is uniformly dis
ID: 1352115 • Letter: T
Question
The seesaw in the figure below is 4.5 m long. Its mass of 20 kg is uniformly distributed. The child on the left end has a mass of 14 kg and is a distance of 1.4 m from the pivot point while a second child of mass 26 kg stands a distance L from the pivot point, keeping the seesaw at rest. (a) Express the conditions for translational and rotational equilibrium for the seesaw. (Assume the seesaw pivots on the support. Use the following as necessary: FR for the force from the child on the right, FP for force from the plank, FL for the force from the child on the left, FS for the force from the support, R for the torque due to the child on the right, P for the torque due to the plank, L for the torque due to the child on the left, and S for the torque due to the support. Indicate the direction with the sign of your answer.) F = = 0 = = 0
(d) Solve to find L, the location of the child on the right. cm
Explanation / Answer
R is negative aND l IS psitive
S is zerom
net torque is = = 0
L = R
FL*x = FR*L
14*9.81*1.4 = 26*9.81*L
L = = 0.75 m