Realizing that she often doesn't have her students' full attention during class, a professor devises an elaborate device on which to stand while she lectures.(Figure 1)The pulley is placed so that the string makes a 45.0-degree angle with the beam. The beam is uniform, 5.00 meters long, and has weightwbwb. The professor stands 2.00 meters from the pivot point and has weightwpwp.

Figure1of1

Part A

Part complete

Find the tensionTTin the rope in terms of the professor's weightwpwpand the weight of the beamwbwb.

Express your answer in terms ofwbwbandwpwp.

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TT=	2.5wb+2wp2+522.5wb+2wp2+52

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Part B

Part complete

If the professor's mass is 60 kilograms and the mass of the beam is 20 kilograms, what is the magnitude of the normal forcennexerted by the beam on the professor? Use 10 meters per second squared for the magnitude of the acceleration due to gravity.

Express your answer in newtons to two significant figures.

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nn=

290

NN

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Correct

Correct answer is shown. Your answer292.9292.9NNwas either rounded differently or used a different number of significant figures than required for this part.

As an incentive to her students, the professor attaches a light basket to the center of mass of the beam and tells them that every student who gets an A on the next test gets to put a ball of mass 4 kilograms into the basket. There are enough students in the class that if a high percentage get A's, the weight of the basket will hoist the professor into the air.

Part C

How many students have to get A's on the test to hoist the professor up into the air?

Express your answer as an integer.

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Hint 1for Part C.How to approach the problem

You could simply add the weights of the balls as another torque in the problem. However, since the basket is located at the center of mass of the beam (where the weight of the beam is applied), the weight in the basket can be considered as an increase in the beam's weightwbwb.

Before you can calculate the weight in the basket that will cause the professor to be hoisted into the air, you have to understand why this should happen at all. Specifically, you need to figure out what condition will lead to her moving upward, instead of staying in equilibrium. Which of the statements below best describes the point at which equilibrium is lost and the professor begins moving upward?

The tensionTTexceeds the normal forceNN.

The tensionTTexceeds the weight of the professorwpwp.

The weight of the beamwbwbexceeds the weight of the professorwpwp.

The weight of the beamwbwbexceeds the tensionTT.

The center of mass of the system moves to the right so that it is no longer supported by the pulley.

The change in weight of the beam causes the tension to be directed at an angle greater than 45 degrees.

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The tension exceeding the weight violates the equilibrium condition on the professor:Fnet=0Fnet=0. If the tension is greater than the professor's weight, then this condition cannot possibly be met, and so she must be accelerting (that is, no longer in equilibrium). While you worked Part A, giving attention explicitly to the equilibrium condition of the beam, the value you used for the normal force was based on the equilibrium condition for the professor, which of course has to be met for the whole system to be in equilbrium.

Now that you know the condition that breaks the equilibrium, you can just set up the equation for that condition and solve it. In this case, setwp=Twp=T, using the expression forTTthat you found above. Solve for the weight of the beam that satisfies this condition. For anywbwbgreater than that value, the professor will be lifted into the air, since the equilibrium conditions for the beamnet=0net=0and for the professorFnet=0Fnet=0can no longer be met.

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