You are not logged in.
Pages: 1
how would you go about solving the following problem
suppose the total tension on the rope is... say 1000N. How do we calculate the acceleration of the man/beam?
what puzzles me is that the man is standing on the beam, and thus the beam is pushing up on the man, but the man is also pulling on the rope which lessens his weight. So i'm not sure how to write the equations of force on each object.
(edit) whoops, i got the masses backwards in the pic. Fixed.
Last edited by mikau (2008-02-09 05:21:04)
A logarithm is just a misspelled algorithm.
Offline
I'm pretty sure that's a stable system (as long as the man is pulling hard enough).
Why did the vector cross the road?
It wanted to be normal.
Offline
what do you mean a stable system? you mean you can treat the man and the beam as a single 75kg object?
A logarithm is just a misspelled algorithm.
Offline
The tension wont be 1000 N. Let T be the tension and a be the acceleration. Then the beam should accelerate upwards and the man downwards. So
From the two equations,
(and ).Last edited by JaneFairfax (2008-02-09 07:53:08)
Offline
Jane, perhaps i didn't make it clear, the idea is that the man is pulling on the rope such that the total tension in the rope is 1000N. So the idea is he IS standing on the beam, but there IS movement. He's pulling his platform upward.
A logarithm is just a misspelled algorithm.
Offline
what do you mean a stable system? you mean you can treat the man and the beam as a single 75kg object?
I meant that nothing will move. I probably should have said that it's in equilibrium. And now I think about it, it doesn't matter how hard he's pulling the rope either, as long as he's holding onto it my statement will be true.
If the man is pulling the platform upward with the rope, then he's also pushing it down with his feet. The reason nothing will move is the same reason that it's impossible to lift yourself up.
Edit: I'm wrong, ignore me.
Why did the vector cross the road?
It wanted to be normal.
Offline
Jane, perhaps i didn't make it clear, the idea is that the man is pulling on the rope such that the total tension in the rope is 1000N. So the idea is he IS standing on the beam, but there IS movement. He's pulling his platform upward.
Well then
I suppose. (Although if Mathsy is right it could be 0 instead.)
Last edited by JaneFairfax (2008-02-09 07:59:45)
Offline
I think the tension in the rope is actually 367.5 N, because
37.5 kg is half the weight of 75 kg, and it has to
be the same on both ends of the rope.
So 37.5 * 9.8 = 367.5 N
But that's if there was no jolt that created the 1000N.
igloo myrtilles fourmis
Offline
mikau wrote:Jane, perhaps i didn't make it clear, the idea is that the man is pulling on the rope such that the total tension in the rope is 1000N. So the idea is he IS standing on the beam, but there IS movement. He's pulling his platform upward.
Well then
I suppose. (Although if Mathsy is right it could be 0 instead.)
I am not really interested in the particular values, my question is how we derive them. How are the forces acting on each other?
Last edited by mikau (2008-02-09 13:09:58)
A logarithm is just a misspelled algorithm.
Offline
In general, if a body of mass M is being lifted vertically upwards by a force F, the resultant of the forces acting on it is F−Mg in the upward direction. By Newtons second law of motion, this is equal to the product of the mass of the body and its acceleration a. Hence
Offline
Pages: 1