Question

W1 has a weight of 220 N acting 0.6m away from the pivot.

How far does W2 whose weight is 180 N need to sit for the see
saw to be in equilibrium.
&
?
W-220 N
W2 - 180N
a) 0.3 m
Ob) 0.73 m
C) 0.5m
O d) 0.83 m

Answer 1

0.73m

Step-by-step explanation:

Given parameters:

W1 = 220N

D1 = 0.6m

W2 = 180N

Unknown

D2 = ?

Solution:

The torque on each side must be balanced;

Torque = force x distance

T₁ = T₂

W1 x D1 = W2 x D2

Insert the parameters and solve for D2;

220 x 0.6 = 180 x D2

132 = 180D2

D2 =
`(132)/(180)` = 0.73m