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Homework Help: Newtonian mechanis

  1. Dec 8, 2004 #1
    I have two questions which I found difficulty in understanding:

    1.) On a isosceles triangular-liked thing, with the base angles equal to 30 degrees, blocks X and Y are put on it on each side, with mass of X = 1kg, and mass of Y = 2kg, S is a spring balance of negligible mass( put on Y's side), and P is a smooth pulley fixed at th top of two smooth inclined planes. What is the reading of S when X is held stationary, and what is the reading when X is moving?

    I have drawn the free-body diagrams of X and Y...There are mgsin 30 acting downwards for both X and Y, and T1 act upwards for X, while T2 act upwards for Y, so I think T1 and T2 should be considered....but seems that no..

    2.) A smooth block of mass 2kg slides down a wedge. The wedge, of mass 10kg, is placed on a horizontal table, and its inclined plane makes an angle of 30 degrees with the horizontal. If the wedge remains stationary all the time, what is the normal reaction of the table acting on the wedge?

    I get the weight and downward component = the reaction, and get 115 N....I wonder what it will be if there is friction? and it is more than 115N but less than 120N. I don't understand why it is this range, and why it is less than 120N

    Please help~ :confused:
  2. jcsd
  3. Dec 8, 2004 #2

    Doc Al

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    Staff: Mentor

    Realize that there is but a single tension throughout the (presumably massless) cord that connects blocks X and Y. So T1 = T2.

    When block X is held fixed, consider the forces on block Y: they must produce equilibrium. But when the blocks are free to slide, block X will accelerate up the plane, while block Y will accelerate down. (Since they are attached, the magnitude of the accelerations is equal.) Consider Newton's 2nd law.

    I presume by "normal reaction of the table" they mean the normal force that the table exerts on the wedge. (There must be a horizontal force component as well, of course, else the block will slide.) So consider the forces on the wedge: The normal force exerted by the sliding block (figure that out), the weight of the wedge, and the force that the table exerts on the wedge. Since the wedge doesn't move, it's in equilibrium. All you need to worry about are the vertical components.

    I'm not sure what your question is here. The normal reaction force will be 115N (if you take g = 10 m/s^2). It would be 120N only if the block didn't slide. As it is, the block accelerates, thus the table doesn't have to support the entire weight of the block.
  4. Dec 8, 2004 #3
    1- Let me ask you a question, what is the spring balance measuring?
    If you understand this you'll see if the tension of the string is needed or not.

    Think about a mass m suspended in a spring balance, what measures the spring balance? Let the balance go, what measures during the free fall?

    2- Consider a block of 10kg and another of 2kg over it. What is the normal "reaction" over the block of 10kg?

    I was thinking in the system block and wedge as a whole, i'll take a second look after reading the post of Doc. It's against my intuition, but science is built to fight the intuition :blushing: .
    Last edited: Dec 9, 2004
  5. Dec 9, 2004 #4
    Thank you for Doc and Evil_kyo's help..I really appreciate it..^-^

    My second question means if the block is not smooth, that is there are friction, what the normal reaction will be?
    How do u get it, Doc? please tell me...

    To Evil_Kyo, you are right, I think I will understand the question better if I can answer your question..^^ I think the spring balance is measuring force needed to hold block X and Y ...but not sure...:frown:
  6. Dec 9, 2004 #5

    Doc Al

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    The normal reaction (of the table on the wedge) depends on the acceleration of the block. If the block slides without friction, the only force that it exerts on the wedge is the normal force; if there is friction between the block and wedge, then that friction force is also exerted on the wedge, which increases the normal reaction. If the friction prevents the block from sliding, then the normal reaction will equal the weight of the wedge plus the block. (In that case, think of the wedge and block as a single object sitting on the table.)

    The spring balance will read the tension in the cord.
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