if an object of 50 Newton is located 2 meter from the pivot point of a uniform 2
ID: 1429835 • Letter: I
Question
if an object of 50 Newton is located 2 meter from the pivot point of a uniform 20 meter seesaw find the distance from the pivot point on an object located on the opposite side that has a weight of 20 newtons. Torque Problems If an object of 50 Newton is located 2 meter from the pivot point of a uniform 20 meter seesaw, find the distance from the pivot point of an object located on the opposite side that has a weight of 20 Newton. 1. A 10 Newton object is placed 2 meters from the pivot point of a uniform 20 meter seesaw while a 15 Newton object is placed 4 meters from the pivot point on the same side. Where would you put a 20 Newton object on the opposite side? 2. A 5 Newton object is placed 3 meters from the pivot point of a uniform 20 meter seesaw while a 15 Newton object is placed 2 meters from the pivot point on the same side. A 10 Newton object sits at 2 meters from the pivot point on the other side. Where would you place a 5 Newton object to balance the seesaw? 3. Be sure to draw a diagram for each problem and indicate the placement of each object for full credit. Call me if you need helpExplanation / Answer
1) when the system is in equilibrium, net torque must be zero.
so, Apply,
Tnet = 0
50*2 - 20*x = 0
==> x = 50*2/20
= 5 m <<<<<<<<<<<---------------------Answer
2) Apply, Tnet = 0
10*2 + 15*4 - 20*x = 0
==> x = (10*2 + 15*4)/20
= 4 m <<<<<<<<<<<---------------------Answer
3) Apply, net torque = 0
5*3 + 15*2 - 10*2 - 5*x = 0
x = (5*3 + 15*2 - 10*2)/5
= 5 m <<<<<<<<<<<---------------------Answer
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