- #1
fterh
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1. A chain of mass M and length L is suspended vertically with its lower end touching a scale. The chain is released and falls onto the scale. What is the reading of the scale when a length of the chain x has fallen? You may neglect the size of the individual links.
2. A smooth wedge, whose central cross-section is a triangle ABC, right-angled at C, rests with the face containing AB on a smooth horizontal plane. When the wedge is held fixed, a particle released from rest, takes a time t1 to slide the full length of CA. The corresponding time for CB is t2. Show that tan A = t2/t1, and find AB in terms of t1 and t2. If the mass of the wedge is n times that of the particle and the wedge is free to move, show that the time of sliding down CA becomes:
3. A light inextensible string passes round a fixed smooth pulley and carries at each end a smooth pulley of mass 1.0 kg. Over each of these pulleys a string hangs, one carrying masses of 1.0 kg and 2.0 kg at its ends, the other masses of 1.0 kg and 3.0 kg. The system is set in motion. Find the acceleration of the pulleys and the tension in the string to which they are attached.
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1. For the first question, I am guessing that in addition to the weight of the portion of the chain lying on the scale, the momentum of the movement of the chain has to be taken into account?
So the weight of the portion would be Mgx/L, while the momentum would be Mx/L*v. v (velocity) = g*t = gx/0.5v. Thus p = Mx/L * gx/0.5v = Mgx^2/0.5Lv.
The answer should be: 3Mgx/L.
2. I got the "Show that tan A = t2/t1" part, and I found AB using pythagoras theorem. However I'm not too sure how to solve for the "free moving wedge" part. I'm guessing that I'd have to calculate the horizontal component of the normal force, to determine the rate of acceleration of the wedge away from the particle. Then I got stuck. Heh.
3. I thought I had it, this is my workings.
http://i.imgur.com/CtSp3.jpg
The answer is: 0.426ms^(-2), 37.5N. My tension is wrong, so I didn't bother calculating for acceleration. I thought my methodology was correct, apparently not. Hmm.
2. A smooth wedge, whose central cross-section is a triangle ABC, right-angled at C, rests with the face containing AB on a smooth horizontal plane. When the wedge is held fixed, a particle released from rest, takes a time t1 to slide the full length of CA. The corresponding time for CB is t2. Show that tan A = t2/t1, and find AB in terms of t1 and t2. If the mass of the wedge is n times that of the particle and the wedge is free to move, show that the time of sliding down CA becomes:
3. A light inextensible string passes round a fixed smooth pulley and carries at each end a smooth pulley of mass 1.0 kg. Over each of these pulleys a string hangs, one carrying masses of 1.0 kg and 2.0 kg at its ends, the other masses of 1.0 kg and 3.0 kg. The system is set in motion. Find the acceleration of the pulleys and the tension in the string to which they are attached.
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1. For the first question, I am guessing that in addition to the weight of the portion of the chain lying on the scale, the momentum of the movement of the chain has to be taken into account?
So the weight of the portion would be Mgx/L, while the momentum would be Mx/L*v. v (velocity) = g*t = gx/0.5v. Thus p = Mx/L * gx/0.5v = Mgx^2/0.5Lv.
The answer should be: 3Mgx/L.
2. I got the "Show that tan A = t2/t1" part, and I found AB using pythagoras theorem. However I'm not too sure how to solve for the "free moving wedge" part. I'm guessing that I'd have to calculate the horizontal component of the normal force, to determine the rate of acceleration of the wedge away from the particle. Then I got stuck. Heh.
3. I thought I had it, this is my workings.
http://i.imgur.com/CtSp3.jpg
The answer is: 0.426ms^(-2), 37.5N. My tension is wrong, so I didn't bother calculating for acceleration. I thought my methodology was correct, apparently not. Hmm.