1. A light plank rests on two scales that read Fg1 = 450 N and Fg2 = 280 N. The scales are separated by a distance of 2.00 m. How far from the woman's feet is her center of gravity?
2.Two window washers, Bob and Joe, are on a 3.00 m long, 380 N scaffold supported by two cables attached to its ends. Bob weighs 765 N and stands 1.00 m from the left end. Two meters from the left end is the 500 N washing equipment. Joe is 0.500 m from the right end and weighs 850 N. Given that the scaffold is in rotational and translational equilibrium, what are the forces on each cable?
If I am missing information let me know. Thanks.
2007-12-15 05:34:12 · 1 answers · asked by loveall 3 in Science & Mathematics ➔ Physics
On the first problem, I imagine that you are asked where the woman is on the plank. Her weight is 450 + 280 = 730 N. We'll put the center of torque where the woman is standing. Then the net torque is
0 = 450 d - 280 ( 2.00 - d )
450 d = 560 - 280d
730 d = 560
d = .76
She is standing .76 meters from the scale reading 450 and 1.24 meters from the scale reading 280.
The second problem is solved the same way. The forces up are the tensions in the cables T1 and T2; the forces down are the weights of the men and the scaffolding.
T1 + T2 = 380N + 765 N + 850 N.
Choose the point where T1 is attached as the center of torque. The net torque about that point is zero
0 = ( T2 ) ( 3.00 m ) - ( 765 N ) ( 1.00 m ) - ( 850 N ) ( 2.50 m ) - ( 380 N ) ( 1.50 m )
Solve the second equation for T2, then use the first to find T1.
2007-12-15 05:43:53 · answer #1 · answered by jgoulden 7 · 0⤊ 0⤋