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fatcat39
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Rotational Equilibrium - Help!
http://photos-a.ak.facebook.com/photos-ak-sf2p/v142/24/75/1238100168/n1238100168_30095452_8017.jpg
A uniform metal bar of mass 100 kg and length L (2.5 m) extends horizontally from a wall and connected to the wall by a pivot (F is exerted on rod by pivot). As seen in the drawing, a 100 kg mass hangs from a massless rope that is L from the building. T, a cable, is attached 0.25 L from the wall. The entire system is in a state of equilibrium.
- Using rotational equilibrium, find tension T
Net Sum of Torques = 0
g = 9.81 m/s^2
So, if you use F as the point to take the torques about, then the bar becomes the moment arm. Therefore, the mass has a force of 981 (M*g) and its torque is 2.5 * 981, or 2452.5. T's torque is 0.25*2.5 * T, right? So, obviously, you would set the two equal to each other. But I think that the bar mass has Mg as well - do i put its torque as 0.5L * 981, or where would I put it?
And are the sigfigs for this problem 1 or 2?
Thanks!
Homework Statement
http://photos-a.ak.facebook.com/photos-ak-sf2p/v142/24/75/1238100168/n1238100168_30095452_8017.jpg
A uniform metal bar of mass 100 kg and length L (2.5 m) extends horizontally from a wall and connected to the wall by a pivot (F is exerted on rod by pivot). As seen in the drawing, a 100 kg mass hangs from a massless rope that is L from the building. T, a cable, is attached 0.25 L from the wall. The entire system is in a state of equilibrium.
- Using rotational equilibrium, find tension T
Homework Equations
Net Sum of Torques = 0
g = 9.81 m/s^2
The Attempt at a Solution
So, if you use F as the point to take the torques about, then the bar becomes the moment arm. Therefore, the mass has a force of 981 (M*g) and its torque is 2.5 * 981, or 2452.5. T's torque is 0.25*2.5 * T, right? So, obviously, you would set the two equal to each other. But I think that the bar mass has Mg as well - do i put its torque as 0.5L * 981, or where would I put it?
And are the sigfigs for this problem 1 or 2?
Thanks!
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