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Hard problem, newtonian mechanics

  • Thread starter mewmew
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  • #1
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Ok, this is a pretty hard problem I dont really know how to start. The set up is as follows, there is a block M and on the top edge of it there is a pulley, set up ontop of mass M is a mass m1, and thats connected to mass m2 that is on the right hand side of mass M. There is no friction and the cords are massless and everything. I really only need help setting up the free body diagrams for the three masses as I should be able to figure it out from there. I am confused also on why exactly mass M accelerates when there is no friction on mass m1 anyways. Any help would be great thanks.
 

Answers and Replies

  • #2
arildno
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Note that since the only external force is gravity, the horizontal position of the system's C.O.M must remain constant.
Since m1 is drawn to the right, M must be drawn to the left.
m1 imparts on the cord a force to the left, this is imparted through the pulley onto M.
Ok?
 
  • #3
ehild
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mewmew said:
I am confused also on why exactly mass M accelerates when there is no friction on mass m1 anyways. Any help would be great thanks.
See the picture. Have I understood your problem well?

There are two crucial points in this problem. One is the acceleration of m1. It is equal to a2 with respect to the big box, which accelerates backwards with "a", so the acceleration of m1 in an inertial frame of reference (with respect to the table) is a1=a2-a.
The other problem is, why does M accelerate at all. Where is the force that makes it accelerate? Well, it is again the tension in the cord. The pulley is pressed by both branches of the cord. The resultant of these two forces from the cord is F. The pulley is attached by a rigid support to the box, that is, the box experiences a force F drawn by red in the picture. The horizontal component of this force is equal to the tension T, that will accelerate the box backwards.

ehild
 
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Yes, the above is correct, thanks alot!
 
  • #5
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Ok, I guess I was wrong, I think I need a little more help, I can't get the f=ma for the masses in the x direction. It seems that Ma=T, m1a1=T, m2a2=T-m2g, with a1=a2. I think it has something to do with a has not only the mass of M but of all three, im not really sure. I guess I am off by quite a bit, but dont know where to start. Thanks alot again
 
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  • #6
ehild
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mewmew said:
a1=a2.
This is wrong, read my post again.

There are two crucial points in this problem. One is the acceleration of m1. It is equal to a2 with respect to the big box, which accelerates backwards with "a", so the acceleration of m1 in an inertial frame of reference (with respect to the table) is a1=a2-a.

ehild
 

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