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Math Help - Physics Question

  1. #1
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    Physics Question

    I just had my last physics lesson before my exam, but I have one question where I disagree with my teacher.

    A dynamics Trolley A of mass 5kg is placed on a horizontal board. It is connected to block B (a hanging mass) of mass 2kg by a light, inextensible string over a frictionless pulley.

    Assuming no friction force on the trolley, calculate the magnitude of the acceleration of the trolley.
    My teacher said use F = ma where F = 2(9.8) and m = 5kg. Which yields the answer 3.92 m/sē

    I decided that the acceleration on the trolley and the hanging mass should be the same. So:

    Trolley:  a = \frac{T}{M_a} (1)where  M_a is the mass of the trolley, T is tension

    Hanging mass:  a = \frac {F}{M_b} where  M_b is the mass of the hanging mass.

     a = \frac{M_b g- T}{M_b} where g is acceleration due to gravity.
    = g -\frac{T}{M_b} (2)

    Equating equations (1) and (2), I get 2.8 m/sē. I'm also aware that using the formula F = ma where m = 7, F = 2g also gives me this answer, but I don't know why. Is my teacher wrong, or am I just over complicating things? Can someone clarify this problem for me?
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  2. #2
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    hi there,

    both methods are correct just that the way you do it is actually the longer way

    what your teacher did was he/she solves the hanging mass first in order to obtain the tension in the string which then allows the usage of Newtons second law.

    What you did was, you are solving both the trolley and the hanging mass at the same time
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  3. #3
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    seems to me your making a mistake

    you have for the trolley:

    acceleration = tension / mass

    and for the hanging mass

    acceleration = force / mass and you rearrange this to be force = mass x gravity

    so for the hanging mass you have force = 2kg x -9.8 m/s^2

    but remember that the force acting on the hanging mass is the same as the force of the tension in the rope for the trolley

    so then you should get

    acceleration of trolley = force of hanging mass / mass of trolley

    look over your equations im not sure how you got the one that reads

    acceleration of trolley = mass hanging x gravity - tension / mass hanging

    using the first two equations i dont see how to get to that one maybe you can show me
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  4. #4
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    if the accelerations are the same then the masses have to be the same

    try the experiment using a 10 kg trolley and a 0.1 kg hanging mass...
    the hanging mass will always have acceleration of 9.8 m/s^2 down but do you think the 10 kg trolly will?
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  5. #5
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    Quote Originally Posted by finch41 View Post
    if the accelerations are the same then the masses have to be the same

    try the experiment using a 10 kg trolley and a 0.1 kg hanging mass...
    the hanging mass will always have acceleration of 9.8 m/s^2 down but do you think the 10 kg trolly will?
    Sorry but you're totally wrong. The acceleration of the hanging mass is defintely NOT g under these conditions. Unless of course the tension in the string is zero .........
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  6. #6
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    sorry
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  7. #7
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    Quote Originally Posted by mr fantastic View Post
    Sorry but you're totally wrong. The acceleration of the hanging mass is defintely NOT g under these conditions. Unless of course the tension in the string is zero .........
    what if friction forces were zero?
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  8. #8
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    Quote Originally Posted by finch41 View Post
    what if friction forces were zero?
    No, not even if friction forces were zero.

    The fact is that the net force on the hanging mass m is (taking downwards as the positive direction) mg - T. Then:

    ma = mg - T \Rightarrow a = \frac{mg - T}{m}.

    a = g only if T = 0. If T = 0 then the string is slack .....

    I suggest you thoroughly review (learn?) the application of Newtonian mechanics to this sort of problem ......
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