- #1
dan987
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I'm currently working on two homework problems that are giving me a bit of trouble. Torque has been a difficult chapter for me. I'm going to worry about just this one for now:
Question 1:
Calculate FA and FB for the beam shown in Fig. 9-49. The downward forces represent the weights of machinery on the beam. Assume the beam is uniform and has a mass of 210 kg.
Fig. 9-49 shows:
http://grab.by/Byt
I know that the net external force must be zero: SUM of Forces = 0
I also know that the net external torque must be zero: SUM of Torque = 0
So I've set up my equation for Torque, where T is Torque:
T = Force x Distance from the Pivot
So if I make my pivot the point at FA, shouldn't I have:
0 = - T(4300N x 2.0M) - T(3100N x 6.0M) - T(2200N x 9.0M) + Tboard + TFB
Solving for the Torque of the Board:
Tboard = 210Kg * 9.8 m/s2 * 5M = 10290
Plugging in all the Torques, 10FB being 10 (distance from FA) x Force (FB)
0 = -8600 - 18600 - 19800 + 10290 + 10FB
When I solve for FB however, and plug that force back into the equation, I'm not getting zero. I'm lost as to where I'm going wrong.
Question 1:
Calculate FA and FB for the beam shown in Fig. 9-49. The downward forces represent the weights of machinery on the beam. Assume the beam is uniform and has a mass of 210 kg.
Fig. 9-49 shows:
http://grab.by/Byt
I know that the net external force must be zero: SUM of Forces = 0
I also know that the net external torque must be zero: SUM of Torque = 0
So I've set up my equation for Torque, where T is Torque:
T = Force x Distance from the Pivot
So if I make my pivot the point at FA, shouldn't I have:
0 = - T(4300N x 2.0M) - T(3100N x 6.0M) - T(2200N x 9.0M) + Tboard + TFB
Solving for the Torque of the Board:
Tboard = 210Kg * 9.8 m/s2 * 5M = 10290
Plugging in all the Torques, 10FB being 10 (distance from FA) x Force (FB)
0 = -8600 - 18600 - 19800 + 10290 + 10FB
When I solve for FB however, and plug that force back into the equation, I'm not getting zero. I'm lost as to where I'm going wrong.
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