Torque diagram problems

  • Thread starter orgo
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  • #1
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I'm stuck on these problems!

1. Suppose a 90.0-g mass is placed at the 82.0 cm mark of a 120.0 g uniform meterstick. At what mark will the meterstick balance?

For this problem, I did TCCW=TCW. I think the TCW is (90 g)*(9.8 m/s^2)*(32 cm). For TCCW, I am not sure because I do not know where the pivot point is.

2. A 10,000 kg bridge of length 10 m is supported at both ends. If a 2000-kg car is parked on the bridge 3.0 m from the left support, what are the supporting forces at the left and right ends?

I drew a diagram for this problem, but I do not know how to calculate the supporting forces.
 

Answers and Replies

  • #2
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set up the momentum balance about the zero mark of the meterstick:
90*9.8*82+120*9.8*50-210*9.8*x=0
210 (=90+120)*9.8 being the reaction force at the pivotpoint
x = the distance of the pivot point from the zero mark
work out x and you have the answer for question 1.
 
  • #3
Doc Al
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bridge problem

orgo said:
2. A 10,000 kg bridge of length 10 m is supported at both ends. If a 2000-kg car is parked on the bridge 3.0 m from the left support, what are the supporting forces at the left and right ends?

I drew a diagram for this problem, but I do not know how to calculate the supporting forces.
Identify all the forces acting on the bridge: its weight, the weight of the car, the left supporting force, and the right supporting force. Draw these on your diagram.

Now apply the conditions for equilibrium:
(1) The forces must balance
(2) The torques about any point must balance

Give it a shot.
 

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