I need help with this problem A see-saw balances with nothing on either side. Th
ID: 1603843 • Letter: I
Question
I need help with this problem
A see-saw balances with nothing on either side. Then two children arrange themselves on the see-saw such that it balances as shown. a. Each child now moves 1 foot farther away from the fulcrum. Will the see-saw remain balanced, tip to the right, or tip to the left? Explain. b. The children return to their original positions. Each child is given a bowling ball to hold. (The bowling balls are identical.) Will the see-saw remain balanced, tip to the right, or tip to the left? Explain.Explanation / Answer
a)
m1 = mass of child on left
r1 = distance of child on left from the fulcrum
m2 = mass of child on right
r2 = distance of child on right from the fulcrum
at equilibrium
m1 r1 = m2 r2
from the diagram ,
r1 < r2
hence m1 > m2
adding 1 ft to each r1 and r2
m1 (r1 + 1) = m2 (r2 + 1)
m1 r1 + m1 = m2 r2 + m2
since m1 r1 = m2 r2 , they cancel out
m1 = m2
which is impossible
hence the see-saw tip to left since m1 > m2
b)
m = mass of bowling ball
(m1 + m) r1 = (m2 + m) r2
m1 r1 + m r1= m2 r2 + m r2
m r1 = m r2
which is impossible
hence the seesaw tip to right
since r2 > r1