Torque has been a difficult chapter

1. Nov 16, 2009

dan987

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.

Last edited by a moderator: Apr 24, 2017
2. Nov 16, 2009

PhanthomJay

Re: Torque

You missed the minus sign on T_board. The torque is in the same direction as the machinery loads torque. The plus or minus sign will bite you every time.

3. Nov 16, 2009

dan987

Re: Torque

Okay, I see what I did there. So then:

57290 = 10FB
so 5729 = FB

edit: I figured it out:
Then I just go to the other side and make the FB pivot at zero, so that:

0 = (2200 x 1M) + (3100 x 4M) + (4300 x 8M) + T_Board - 10FA
0 = 2200 + 12400 + 34400 + 10290 - 10FA
FA = 5929

Thanks for the help!

Last edited: Nov 16, 2009