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Motion of connected masses

  1. Aug 15, 2015 #1
    1. The problem statement, all variables and given/known data
    Two scale-pans each of mass 3 lbs are connected by a string passing over a smooth pulley. Show how to divide a mass of 12 lbs between the two scale-pans so that the heavier may descend a distance of 50 ft in the first 5 seconds.

    2. Relevant equations
    F = ma, mg - T = ma, where, T is the tension on the string, (acceleration of the mass on one end of the string) = -(acceleration of the mass on the other end)

    3. The attempt at a solution
    well, in my attempt to this question, I considered the pan and the divided mass at one of the ends as a connected system of masses having a common acceleration, say, 'a' and assumed the divided masses as x and (12-x) where (12-x) is the heavier mass. Then by Newton's second law I came up with the equation : at one end, (3+12-x)g - T = (3+12-x)a and similarly for the other end. Here, I considered the downward direction as positive. Using the given data I got a = 4 ft/s^2. Anyway, after solving the force equations for both the ends and using the value of a, I got x =12/5 lbs and 12-x = 48/5 lbs. But the answer is given to be 57/8 lbs and 39/8 lbs. Please help me out.
     
  2. jcsd
  3. Aug 15, 2015 #2

    ehild

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    Welcome to PF!

    Show your work in detail. You might have miscalculated something.
     
  4. Aug 15, 2015 #3
    yeah the solution goes like this: at one end, let us assume the heavier mass to be (12-x) lbs and at the other end the lighter mass be x lbs. the mass of the scale-pan on both the ends is 3 lbs. let us assume the downward direction as positive. let T be the tension on the string and g = 32 ft/s^2. also let the common acceleration of the system be a ft/s^2. Now at one end let us consider the mass of the pan and that of the divided mass (12-x) lbs as one single mass. so I suppose that we may neglect the action-reaction forces between them. here according to the question, the tension T acts only on the pan. therefore by Newton's second law, the force equation for this heavier end is: {3+(12-x)}g - T = {3 + (12-x)}a ---- (1), since at this end acceleration 'a' acts downward. In a same way the force equation for the other end is: (3+x)g - T = -(3+x)a ------ (2), as here 'a' acts upward. Now assuming the system to be initially at rest, the heavier mass descends 50 ft with acceleration a ft/s^2 in 5 seconds i.e. 50 = 0*5 + 1/2*a*(5)^2 implying a = 4 ft/s^2. Now solving equations (1) and (2) and using the value of 'a' we get x = 12/5 lbs i.e. the mass of the lighter part and (12-x) = 48/5 lbs i.e. the mass of the heavier part. But the given answer is something else. I need some urgent clarification.please help.thank you.
     
    Last edited: Aug 15, 2015
  5. Aug 15, 2015 #4

    ehild

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    Your equations are correct, but you did something wrong afterwards. You should have shown all your derivation.
     
  6. Aug 16, 2015 #5
    yes, you were right. I considered g = 10 m/s^2. It was very silly of me. Anyway, thanks a lot for helping me out.
     
  7. Aug 16, 2015 #6
    By the way, will you please tell me why in case of considering two masses as one single mass, we neglect the internal forces between them (like tension, action-reaction forces). also, which one is preferable, to consider the masses individually and analyse them or to see consider them as a whole.
     
  8. Aug 16, 2015 #7

    ehild

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    The equation which is valid for the combined masses and their common acceleration is derived from the equations for the individual masses, by eliminating the internal forces (tension, here). I suggest to handle all bodies separately.
     
  9. Aug 16, 2015 #8
  10. Aug 16, 2015 #9
    when we refer to earth as a reference frame do we mean " surface of the earth", or some coordinate system passing through the centre of the earth.
     
  11. Aug 16, 2015 #10

    ehild

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    It depends. If you investigate the motion of bodies on or near to the Earth surface, the frame of reference is the surface of the Earth, considered as an inertial frame of reference usually. When you deal with the motion of satellites or the Moon, you use the centre of the Earth as reference.
     
  12. Aug 16, 2015 #11
    In the cases when an object is connected to a string at one end, does the string exert tension on the object even when no force is applied at the other end of the string? also if no resistive forces like friction, pull or any other force of that kind acts on the object opposite to the direction of string, will the string still exert tension on the object? According to me tension acts on the object as a result of a pull on the string by the object i.e. it is a reaction to the pull of the object on the string when the two are in contact (connected). But is it the case really?
     
    Last edited: Aug 16, 2015
  13. Aug 16, 2015 #12

    ehild

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    If you connect an object to a string and the other end is not connected to anything, then the string is slack. There is no tension, no force on the object exerted by the string.
    You need to connect both ends of a string to something in order to make it taut and get tension in it.
    In case of a massless string, its tension balances the forces exerted on the object to keep it fixed or give it a certain acceleration. So you are right, the tension is the result of a pull on the string by the object and also the pull by an other object at the other end. In the massless string, the tension is the same everywhere. It is the same at the other end, but that force of tension must act on something which pulls it back, so as the net force on the string is zero. Otherwise it would have infinite acceleration.
     
  14. Aug 16, 2015 #13
    then if we consider a case where suppose a) two bodies are connected by a string and they are resting on a frictionless surface b) an one end of the string some force is applied while at the other end no external pulling force is applied c) the object in this end is found to be accelerating in the direction of the force applied on the object on the other end and a tension also acts on it. What is the pulling force on this object on which no external force is applied initially but give rise to a tension on the string?Is it only the result of the force of action of the object on the string as they are in contact?
     
    Last edited: Aug 16, 2015
  15. Aug 16, 2015 #14

    ehild

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    If two objects are connected by a string, and one is pulled by an external force F, the acceleration a of this object is determined by the pulling force F - tension T : m1a=F-T.
    Only the tension acts on the object on the other end. It accelerates, and the accelerations of both objects are the same: m2a = T.
    You can add the two equations, getting (m1+m2)a=F. a=F/(m1+m2). Knowing a, you get the tension as ##T=\frac{m_2}{m_1+m_2}F##
     
  16. Aug 16, 2015 #15
    got it. thanks!
     
  17. Aug 17, 2015 #16
    Sorry, I think I still have a confusion. In our previous case the object on the other end accelerates only because of the tension on the string. but how this tension itself is generated. as because tension on the object is due to some force on it acting in the opposite direction. but here no such force is acting on it. Is it rather the pull of the object on the string (action) the result of which is the tension of the string on the object (reaction)?
     
  18. Aug 17, 2015 #17

    andrewkirk

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    It can depend on what seems most intuitive to you. Personally, I find it most intuitive to regard this as a problem of a single mass moving in one dimension. THe dimension is the line of the string from one pan up to the pulley, around the pulley and down the other side. All motion is along that line, so it's a 1D problem and we can consider the system as a single mass moving in 1D. If the pans were swaying like pendulums it would be 2D and if they were swinging around it could be 3D, and we'd probably have to consider indivisual masses, but they're not, so it's easy. We just need to arbitrarily choose which direction along the string is positive. To me that is most naturally the direction of motion, which is pointing up to the pulley from the lighter pan and then down from the pulley to the heavier pan.

    To get the acceleration on the combined mass in that 1D space you still need to consider the gravitational forces on the two pans though. The difference between the two forces is the force on the combined mass. When you approach the problem this way you don't need to consider the tension in the string.
     
  19. Aug 17, 2015 #18

    ehild

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    You have force on the object and that force has the magnitude equal to the tension in the string. According to Newton's Third Law, the object also exerts force on the string and that is equal to the tension.

    You can imagine the string as a spring. If you connect the spring to a wall, and pull the other end, tension will be generated in the spring. Both you and the wall exert force on the spring. Also, the string pulls your hand with a force equal to the tension, and it also pulls the wall with a force of the same magnitude.

    If you connect an object at one end and you pull the other end of the spring, first the object is in rest, and you stretch the spring. That stretching causes tension, the tension travels along the spring and reaches the object at the other end, making it accelerate.

    A string has some elasticity, but the spring constant is very large, so the change of length is negligible.
    Do you know Slinky? Play with it. You will understand a lot about how tension and force are built up in an elastic body.
     
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