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Math Help - Couple of general/abstracted mechanics questions

  1. #1
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    Couple of general/abstracted mechanics questions

    Hi I have two questions that I'm stuck on. I don't want solutions but any hints/pointers would be much appreciated.



    1) An object is projected vertically upwards with a velocity of u m/s, and after t s another object is projected upwards from the same point with the same velocity. Find, in terms of u, t and g the time which elapses between the second object's projection and the collision between the objects.

    At first I tried finding the times when the displacements of equal, using (T+t) and t for A and B respectively. Then got nowhere...


    2) A light string of length a is attached to two points A and B on the same level and a distance b apart, where b<a. A smooth ring of weight W is threaded on the string and is pulled by a horizontal force, P, so that it rests in equilibrium vertically below B. Show that the tension in the string is

    [W(a^2 + b^2)]/(2a^2)

    No idea on this one...any hints?
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  2. #2
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    Re: Couple of general/abstracted mechanics questions

    For (1) you are on the right track. Where did it fail? You should have equations:

     u(t+T)-\frac 1 2 g(t+T)^2 = ut - \frac 1 2 g t^2

    You can then solve for t.

    For (2) if you draw the free body diagram you have forces in the vertical direction:  T \sin \theta + T = W where  \theta is the angle of the string to the first support. In the horizontal direction you have  T \cos \theta = P. And from the length of the string being A you have  \frac B {\cos \theta} + B \tan \theta = A. From this you have enough to get the final expression, but it takes some playing around. I suggest you get this last expression into terms of  \cos \theta.
    Thanks from kinhew93
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    Re: Couple of general/abstracted mechanics questions

    Quote Originally Posted by ebaines View Post

    For (2) if you draw the free body diagram you have forces in the vertical direction:  T \sin \theta + T = W where  \theta is the angle of the string to the first support. In the horizontal direction you have  T \cos \theta = P. And from the length of the string being A you have  \frac B {\cos \theta} + B \tan \theta = A. From this you have enough to get the final expression, but it takes some playing around. I suggest you get this last expression into terms of  \cos \theta.
    Ok all that is clear to me, but I can't seem to get an equation for T without P or any trig functions...What is the next step?

    thanks
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    Re: Couple of general/abstracted mechanics questions

    Quote Originally Posted by ebaines View Post
    For (1) you are on the right track. Where did it fail? You should have equations:

     u(t+T)-\frac 1 2 g(t+T)^2 = ut - \frac 1 2 g t^2

    You can then solve for t.

    For (2) if you draw the free body diagram you have forces in the vertical direction:  T \sin \theta + T = W where  \theta is the angle of the string to the first support. In the horizontal direction you have  T \cos \theta = P. And from the length of the string being A you have  \frac B {\cos \theta} + B \tan \theta = A. From this you have enough to get the final expression, but it takes some playing around. I suggest you get this last expression into terms of  \cos \theta.
    WRONG APPROACH!!!!
    Solving the above suggested equation yields t=(5T^2)/(u-10T) which is wrong!!!

    Kinhew...

    The first object needs (u/10 ) seconds to reach the max height and then it stops for an instant before it starts to fall free down.
    At this moment the second object is at a height H travelling (u/10 -t) seconds because the delay is t seconds.
    The height H can be calculated easily from the formula of the Uniform Decelarated Motion with g =10 and it is H = (u^2/20)-5t^2
    the gap between the two object at this moment A at the max height and B at a height H is H = 5t^2.
    The two objects will continue the motion but now the object A is falling down with U A. Motion while the onbect B continues the U.D M upwards.
    They will meet at a point P x seconds after.
    Get the UAM and UDM formulas (put x as time)and add them together and equate them with 5t^2. solve the equation to get x = t/2 !!! this is the corect solution
    therefore the total time needed for the coallision is u/10+t/2
    EXample: If u=40 m/s and t =1 second then they will meet at a height of 78.75 m after 4.5 seconds the first object was launched.
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  5. #5
    MHF Contributor ebaines's Avatar
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    Re: Couple of general/abstracted mechanics questions

    Quote Originally Posted by MINOANMAN View Post
    WRONG APPROACH!!!!
    Solving the above suggested equation yields t=(5T^2)/(u-10T) which is wrong!!!

    The approach I suggested is correct. Check your math, my approach it yields t = u/g - T/2. I think the difference is that I define t=0 when the second projectile is fired, not the first.
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