# Newton's Laws of motion!

1. Jan 23, 2009

### psykatic

1. The problem statement, all variables and given/known data

A smooth ring A of mass m can slide on a fixed horizontal rod. A string tied to the ring passes over a fixed pulley B and carries a block C of Mass M (=2m) as shown in the given figure. At a instant at the string between the ring and the pulley makes an angle $$\theta$$ with the rod, (a)Show that, if the ring slides with a speed v, the block descends with speed v cos$$\theta$$. (b) With what acceleration will the ring start moving if the system is released from rest with $$\theta=~30^\circ$$?

2. Relevant equations

Newtons Equations, free body diagram

3. The attempt at a solution

Well, this question happens to be from a textbook. And its a solved one too.. The solution which they have given is quite complicated! I thought I'd get a brief explanation of "why and how" over it, and yes an alternative method would be highly appreciated!

The solution, is like this (as given in the book),

Suppose in a small time interval $$\delta t$$ the ring is displaced from A to A' and the block from C to C'. Drop a perpendicular A'P from A' to AB. For small displacements A'B~ PB (I didnt get this!), since the lenght is constant (?????), we have,
AB+BC= A'B+BC'
AP+PB+BC=A'B+BC'
AP=BC'-BC=CC' (as A'B=PB)

AA'$$cos\theta$$= CC'

or $$\frac{AA'cos\theta}{\delta t}=~\frac{CC'}{\delta t}$$

Therefor, (velocity of the ring)$$cos\theta$$= (velocity of the block)

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• ###### freebody.bmp
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Last edited: Jan 23, 2009
2. Jan 23, 2009

### rl.bhat

When the ring moves with a velocity v horizontally, its component along the string is v*cos(theta). Since the length of string between the ring and block is costant, the velocity of the block is equal to v*cos(theta).

3. Jan 23, 2009

### psykatic

Well, did you have a look at the diagram? It hasnt been approved yet..

4. Jan 23, 2009

### rl.bhat

Suppose in a small time interval LaTeX Code: \\delta t the ring is displaced from A to A' and the block from C to C'. Drop a perpendicular A'P from A' to AB. For small displacements A'B~ PB (I didnt get this!), since the lenght is constant (?????), we have,

When A and A(dash) are very close, AB and A(dash)B are nearly equal. If you take BA(dash) equal to BP, AP becomes A*A(dash)cos(theta) and A*A(dash)/t = velocity. The length of the string is ABC = A(dash)BC(dash)

5. Jan 24, 2009

### psykatic

okay, thats fine with me, I've several more problems over the free body diagrams! I'll post it in mean time! Thank You!