Is Your Calculation of Net Torque Correct?

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The discussion focuses on calculating net torque using the formula T = F(d)sin(x). A participant initially calculated the net torque as -994.2 Nm but received feedback on the distances used for the forces. It was clarified that the second term should be based on a distance of 6m from the pivot, not 4m, and the bottom force of 150N is actually 15m from the pivot. The importance of using the perpendicular distance to the pivot for torque calculations was emphasized. This clarification helped resolve confusion regarding the correct distances and angles involved in the torque calculation.
Intrusionv2
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Homework Statement



Find net torque.
2dbqwl0.jpg


Homework Equations

T = F(d) and T = F(d)sinx

Torque = T
Force = F
d = distance

The Attempt at a Solution



100(0) - 200(4)sin40 - 120(14) + 150(8) = -994.2 Nm

Did I do this correctly? Notice I crossed out the 400N also because I think that it does not affect it.
 
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Welcome to PF.

I think your second term is 6m from the pivot not 4m.

Your bottom force - 150N - is 15m from the pivot not 8m as your equation suggests.
 
Ah, thanks, but how is 150N 15m away from the pivot point? Do I just go up? Sorry I do not understand :(
 
Intrusionv2 said:
Ah, thanks, but how is 150N 15m away from the pivot point? Do I just go up? Sorry I do not understand :(

It's the projection of the force to a line that is perpendicular to the pivot. That is the way you should be taking Torque.

For instance the one at an angle of 40° if you extended that out you would see that the lever arm that is perpendicular to the force is Sin40° times the 6m length.

For the bottom force it is acting horizontally, but it is acting through the vertical line from the pivot at a distance of 14 + 1 = 15.
 
Ahhh...that clears up so many things. Thank you so much!
 
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