Question
1. [1pt] A professor of mass m = 65 kg is seated on a light harness connected to a rope and pulley system as shown. The...
1. [1pt]
A professor of mass m = 65 kg is seated on a light harness connected to a rope and pulley system as shown. The professor's feet touch a uniform plank of mass mplank = 20 kg which is supported by a hinge at the wall. A mass M in a light bucket is suspended from the right end of the plank. The professor is a distance ℓ = 2.90 m from the hinge; the plank has length L = 6.20 m.
- Are you kidding? This situation cannot be in equilibrium for any values of M.
- If the plank remains horizontal, the concept of static equilibrium can be used to calculate the tension in the rope.
- The tension in the rope leads to a torque of magnitude TL on the plank, taken about the hinge.
- If M = 0 and the plank remains horizontal, the tension in the rope does not depend on the ratio of ℓ to L.
- If the plank remains horizontal, the force of the professor on the plank is equal to the tension in the rope.
Answer true or false to the statements below; e.g., if the first one is true and the rest are false, enter TFFFF.
Answer:
2. [1pt]
For the values given above, find the tension in the rope if M is zero.
Answer:
3. [1pt]
What is the minimum value of M needed to make the professor just lose contact with the plank?
T m p lank L
Answers
Solution: 1) FTTTT From the expression of tension, T=- L+1 (65kg)(9.8m/s?)(2.90m)+(0+20kg 6.20m)(9.8m/s?) 6.20m +2.90 m = 269.76 N N= L+1 0 L+1 M=m-" = 65 kg – 20 kg = 55 kg
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