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Equilibruim: pivot points

by mandy9008
Tags: equilibruim, pivot, points
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Jul7-10, 11:26 AM
P: 127
1. The problem statement, all variables and given/known data
A beam resting on two pivots has a length of L = 6.00 m and mass M = 72.0 kg. The pivot under the left end exerts a normal force n1 on the beam, and the second pivot placed a distance l = 4.00 m from the left end exerts a normal force n2. A woman of mass m = 51.1 kg steps onto the left end of the beam and begins walking to the right as in the figure below. The goal is to find the woman's position when the beam begins to tip.


a. What is n1 when the beam is about to tip?
b. Use the force equation of equilibrium to find the value of n2 when the beam is about to tip.
c. Using the result of part (a) and the torque equilibrium equation, with torques computed around the second pivot point, find the woman's position when the beam is about to tip.

3. The attempt at a solution
a. n1=0 N
b. n2=mg
n2= (51.1kg)(9.8 m/s2)
n2=500.78 N
c. since F is 0 in part a, I would think that this answer will be zero as well
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Jul7-10, 02:31 PM
P: 1,261
You have to include the mass of the beam itself as-well.
Jul7-10, 03:02 PM
P: 127
okay, so n2 will be 1206.38 instead, assuming that you meant: n2= (51.1kg + 72.0 kg)(9.8 m/s2). so what do i need to do for part c?

Jul7-10, 03:12 PM
P: 1,261
Equilibruim: pivot points

That looks about right. Whats the equation for torque? And the instant before it begins to tip, what is the net torque? It might help if you draw a picture of this situation, consider what torques are in what direction, remember to include the beam.

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