Two people carry a heavy electric motor by placing it on a light board of length 1.60 m. One person lifts at one end with a force of 410 N, and the other lifts the opposite end with a force of 630 N.
What is the weight of the motor?
Where along the board is its center of gravity located?
Express your answer as a distance measured from the end where the 410 N force is applied
2006-11-07 19:17:09 · 4 answers · asked by Theresa C 2 in Science & Mathematics ➔ Physics
630 N + 410 N = 1040 mg
m = 1040 N / g = 106 kg
410 * x = 630 * (1.6 - x) = 1008 - 630 x
1040 x = 1008
x = 0.96923 m
2006-11-07 19:23:18 · answer #1 · answered by feanor 7 · 0⤊ 0⤋
If the board is not travellig up or down, all the vertical forces must add up to zero. Therefore the weight of the motor is (410+630) N. In order for the board not to rotate, the torque from all forces must add up to zero. The torque is measured around the center of gravity. Take that location to be a distance s from the 410N force. Then the torque from that force is 410*s. The torque from the 630N force is 630*(1.6 - s). If the board stays horizontal, the weight is placed at the center of gravity and does not contribute to torque. Therefore 410*s = 630(1.6-s), 410*s = 1.6*630 - 630*s (410+630)*s = 1.6*630 and s = 1.6*630/(410+630).
2006-11-07 19:32:09 · answer #2 · answered by gp4rts 7 · 0⤊ 2⤋
the weight of the motor is (410+630)N = 1040N
the mass of the motor is 1040/9.81=106.01kg
if you take the sum of the moments around the 410N end and set them equal to zero (meaning there is no moment, which means that the board would not rotate) then you have...
-1040x+630(1.6)=0
630(1.6)=1040x
x=(630)(1.6)/(1040)
x=.9692 meters from the 410N end
2006-11-07 19:24:00 · answer #3 · answered by igot4onit 2 · 1⤊ 0⤋
(410+630)/9.8=106.12kg
Pivot at person with 410N
d*(410+630)=1.6*630
d=0.97m from 410N
2006-11-07 19:22:31 · answer #4 · answered by Anonymous · 1⤊ 0⤋