A uniform aluminum beam 9.00 m long, weighing 273 N, rests symmetrically on two

Question

A uniform aluminum beam 9.00 m long, weighing 273 N, rests symmetrically on two supports 4.95 m apart (see figure). A boy weighing 648 N starts at point A and walks toward the right.

(a) In the same diagram construct two graphs showing the upward forces FA and FB exerted on the beam at points A and B as functions of the coordinate x of the boy. Let 1 cm = 100 N vertically, and let 1 cm = 1.00 m horizontally. (Do this on paper. Your instructor may ask you to turn in this work.)

(b) From your diagram, how far beyond point B can the boy walk before the beam tips? m

(c) How far from the right end of the beam should support B be placed so that the boy can walk just to the end of the beam without causing it to tip? m

Explanation / Answer

Assumption # 1: The 9m beam is center along the 5m-wide supports, i.e. support A is 2m from the left edge and support B is 2m from the right edge.
2) Beam is 9.0m9.0m long and there is 5.0m5.0m between the supports, which gives us
9-4.95/2 = 2.025 mon each side of the beam beyond the supports.


Part a)Part a)

Beam is in equilibrium, and its center of the mass is located 4.5m from each end of the beam. This means that the center is located 2.48m from each support.(4.5-2.05)
Now we can set up an equation for the first part. We want forces on each side of the support BB to be equal.
On one side we have 2.48m distance to BB and weight of 273N, and on another side we have distance xx and a weight of a boy, 648N

Equilibrium equation:

2.48m * 273N=x648N

x= 1.044 m

This means that the boy can walk 1.044m beyond point BB before the beam tips.

We want to move support BB to the right for xx meters. Remember that it is 4.5m from the end of the beam to its center.
Right now, support BB is 2.025 meters from the right end and 2.48 meters from the center of the beam. Now let xx represent the new distance from the end of the beam to the support B.

Then, B is xx meters from the right end and (4.5x)(4.5x) meters from the center of the beam.
This time we want the boy to be able to walk to the right end of the beam. Set up the equilibrium equation:

(4.5mx) * 273N= x * 648 N

1228.5 N - 273x = 648x

1228.5=921x

x = 1.33 m

Thus, support BB should be placed 1.33 meters from the right end so that the boy can walk just to the end of the beam.

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