contestada


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

Respuesta :

quasarJose

Answer:

0.73m

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 = [tex]\frac{132}{180}[/tex]   = 0.73m