- #1
Kavorka
- 95
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I'm having trouble with two review questions:
1) Julie has been hired to help pain the trim of a building, but she is not convinced of the safety of the apparatus. A 5 m plank is suspended horizontally from the top of the building by ropes attached at each end. Julie knows from previous experience that the ropes being used will break if the tension exceeds 1 kN. Her 80 kg boss dismisses Julie's worries and begins painting while standing 1 m from the end of the plank. If Julie's mass is 60 kg and the plank has a mass of 20 kg, over what range of positions can Julie stand to join her boss without causing the ropes to break?
I believe you need a summation of forces and torques set to zero, and then solve for Julie's distance in the torque equation. My confusion is how to draw the free-body diagram. I have the two tensions as T/2 (I believe they mean the total tension to be 1kN) with the ropes horizontal, but what vertical force balances the weight of the plank and people?
2) The position vector of a particle whose mass is 3 kg is given by r = (4 + 3t^2)j, where r is in meters and t is in seconds. Determine the angular momentum and net torque about the origin acting on the particle.
I found the velocity vector dr/dt = 6tj and the acceleration vector dv/dr = 6j
Torque is position cross force, and the force vector F=ma is 18j. Angular momentum is position cross linear momentum, and the linear momentum vector is 18tj.
My problem is, if you cross all these j vectors don't they equal 0? Wouldn't both the net torque and angular momentum be 0? It seems this kind of problem should have a real answer, so I'm very much doubting myself.
1) Julie has been hired to help pain the trim of a building, but she is not convinced of the safety of the apparatus. A 5 m plank is suspended horizontally from the top of the building by ropes attached at each end. Julie knows from previous experience that the ropes being used will break if the tension exceeds 1 kN. Her 80 kg boss dismisses Julie's worries and begins painting while standing 1 m from the end of the plank. If Julie's mass is 60 kg and the plank has a mass of 20 kg, over what range of positions can Julie stand to join her boss without causing the ropes to break?
I believe you need a summation of forces and torques set to zero, and then solve for Julie's distance in the torque equation. My confusion is how to draw the free-body diagram. I have the two tensions as T/2 (I believe they mean the total tension to be 1kN) with the ropes horizontal, but what vertical force balances the weight of the plank and people?
2) The position vector of a particle whose mass is 3 kg is given by r = (4 + 3t^2)j, where r is in meters and t is in seconds. Determine the angular momentum and net torque about the origin acting on the particle.
I found the velocity vector dr/dt = 6tj and the acceleration vector dv/dr = 6j
Torque is position cross force, and the force vector F=ma is 18j. Angular momentum is position cross linear momentum, and the linear momentum vector is 18tj.
My problem is, if you cross all these j vectors don't they equal 0? Wouldn't both the net torque and angular momentum be 0? It seems this kind of problem should have a real answer, so I'm very much doubting myself.
