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Taking components of a system containing multiple vectors.

Q.In the arrangement shown in fig. the ends P and Q of an inextensible string move downwards with uniform speed u. Pulleys A and B are fixed. The mass M moves upwards with a speed.

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ANSWER PROVIDED BY MY TEXT

MrRRkou.png


In the diagram ##vcos\theta=u## therefore ##v=u/cos\theta##

HOW I TRIED TO DO THE PROBLEM:

4jvsyA6.png


Here both the ##usin\theta## cancel out but the component ##ucos\theta## of both the vectors u add up. Therefore ##2ucos\theta=v## or ##v=2ucos\theta##

But this is a completely different answer regarding the former one, and the latter answer is marked wrong and the former correct. Why can't we add up the vectors like the latter case?


 
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CWatters

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The first is a velocity diagram and second is a force diagram.

Consider the first diagram... Suppose you add a car moving horizontally under the mass. To show the relative velocity between mass and car you would draw a vector arrow between the two. Adding that arrow to the diagram doesn't change the velocity of the mass or car does it?
 
CWatters - Thanks for your reply.
Does it mean that in this case ##ucos\theta## is just the vertical component of the velocity u and has nothing to do with or doesn't contribute anything to v?

And, what about ##vsin\theta## ? Which entity provides this velocity? Where is it? And what is happening to it?

It will be really helpful if you can provide an explanation. Thanks
 
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CWatters

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CWatters - Thanks for your reply.
Does it mean that in this case ##ucos\theta## is just the vertical component of the velocity u and has nothing to do with or doesn't contribute anything to v?
Which case?

ucos(theta) only makes sense if you are talking about the force diagram. So it's got nothing to do with velocities.

I think you are getting confused because both cases appear to use similar trigonometry but in fact they are quite different. First thing to note is that the "right angled triangle" is not the same in the two cases...

MrRRkou.png 4jvsyA6.png

Looking at the left hand (velocity drawing)....

VCos(theta) is the component of V acting in the direction of the rope (so it is the velocity of the rope = u)
VSin(theta) is the component of V acting at right angles to the rope.

Looking at the right hand drawing (force diagram)...

UCos(theta) is the vertical component of the force U.
USin(theta) is the horizontal component of the force U.
 
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I think their diagram is confusing too. The points of the rope do not move with that velocity. It is obvious for the point in the middle of the rope which has no horizontal component of velocity. Even though the result seems to be OK, I cannot see how to justify the "method".

If the ends move down by u*dt, the rope in the middle should get shorter by the 2 u*dt (or each half by u*dt).
You have to write a relationship between the shortening of the rope in the middle and the vertical displacement of the middle point. This will provide the relationship between the two velocities.
 
Why can't we take components of velocity here?
 

haruspex

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Why can't we take components of velocity here?
You can; that is the textbook method, and the method described in post #4.
The tricky part is realising that you need to take the component of v in the direction towards the pulley, not the component of u (as speed of the angled sections of strings) in the direction of the movement of the mass.
 
I am not able to understand why the method that the questioner has used is wrong please explain
 
I am not able to understand why the method that the questioner has used is wrong please explain
Since the final motion of the body is vertically upwards,the magnitude of velocity at that direction must be it's net magnitude.
This means that the velocity of the strings must be components of that velocity.
 

haruspex

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I am not able to understand why the method that the questioner has used is wrong please explain
The second method has two independent flaws.
First, it supposes that the piece of string immediately adjacent to the mass is moving directly towards the pulley (at speed u). As has been noted, it is moving vertically upwards at the same speed as the mass.
Its total speed must be more than u. It is getting closer to the pulley at rate u, but it also has a tangential velocity so as to result in a net upward velocity. Thus v> u.

The second error is adding the two velocities, for P and Q, thus calculated. By symmetry, the value of v calculated to make it all work on one side also works for the other side.
 

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