Mechanics, linear motion help

craig · #1 $Old$ January 9th, 2009, 01:42 PM

Hi, this is probably something basic I am missing but I can't get my head around it.

$mechanics-linear-motion-help-tension-question.bmp$

In the question you are told that A and B are particles of mass 3m and m respectively, each attached to a light inextensionable string of length 3h. The string passes over a smooth pulley attached to a rough table. The coefficient of friction between the table and A is $u$ .

The system is released from rest and begins to move...

1. Show that until B hits the floor, the acceleration of A is $\frac{1}{4}(1-3u)g$ , this I was able to do.

2. Find the speed of A immediatly before B hits the floor, managed to do this as well, got it to be $\sqrt{\frac{gh(1-3u)}{2}}$

3. After B hits the floor, A continues to move, coming to rest before reaching the pulley. Deduce the range of values of $u$ for the motion above to be possible.

It is the last bit I can't get my head round, any ideas?

Thanks in advance

Craig

HallsofIvy · #2 $Old$ January 10th, 2009, 05:47 AM

I answered (a) then saw that you had already done that! Very good.

For (c), a "range" of values for $\mu$ must have both an upper and lower bound. The upper bound is easy: $\mu$ cannot be so large as to have no motion at all. The total force on the system is $mg(3\mu- 1)$ (you must have calculated that when you did (a)) and the masses will not move if that is negative $mg(3\mu-1)< 0$ leads to $\mu< 1/3$ . Now you are told that the mass A, which, as you calculated for (b) has speed $\sqrt{\frac{gh(1-3u)}{2}}$ when B stops falling and so the only force on mass A is the friction force $3mg\mu$ . The acceleration due to that is $3g\mu$ and that must stop A before it reaches the pulley.

A is initially distance 2h from the pulley. After B has fallen a distance h, A will have moved distance h also and so will be distance h from the pulley. Using decceleration $3g\mu$ to determine the time required for A to stop. Calculate the distance A will have traveled. Find the value of $\mu$ so that distance would be exactly h. That is the lower bound for $\mu$ .

craig · #3 $Old$ January 10th, 2009, 02:42 PM

Quote:

Originally Posted by HallsofIvy $View Post$

I answered (a) then saw that you had already done that! Very good.

For (c), a "range" of values for $\mu$ must have both an upper and lower bound. The upper bound is easy: $\mu$ cannot be so large as to have no motion at all. The total force on the system is $mg(3\mu- 1)$ (you must have calculated that when you did (a)) and the masses will not move if that is negative $mg(3\mu-1)< 0$ leads to $\mu< 1/3$ . Now you are told that the mass A, which, as you calculated for (b) has speed $\sqrt{\frac{gh(1-3u)}{2}}$ when B stops falling and so the only force on mass A is the friction force $3mg\mu$ . The acceleration due to that is $3g\mu$ and that must stop A before it reaches the pulley.

A is initially distance 2h from the pulley. After B has fallen a distance h, A will have moved distance h also and so will be distance h from the pulley. Using decceleration $3g\mu$ to determine the time required for A to stop. Calculate the distance A will have traveled. Find the value of $\mu$ so that distance would be exactly h. That is the lower bound for $\mu$ .

Thank you

, I don't know why but with all the unknown quantities I just could not get my head around the maths

Thanks again