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System of moving bodies

  • Thread starter undertow2005
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  • #1
undertow2005
i'm working without my textbook and going by the some solutions that i have so i need some help

1) Two masses, m1 and m2, joined with a cord are dragged along the horizontal plane surface. the applied force is F=20N, mass m1=m2=3kg, and the acceleration is 0.50 m/s. what is the tension, T, in the connecting cord if the frictional forces on the two blocks blocks are equal? How large is the frictional force, F, on each block?
diagram:
| m1 |-----| m2|------>F
 

Answers and Replies

  • #2
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can you explain what you have done to try and solve this?
 
  • #3
undertow2005
i've solved simpled problems--problems with only one body moving. so i tried to apply those principles here:

for tension:
T=3kg*.50
i know know that is wrong since there is a frictional force. i just dont know how to compute it since the coefficient of friction is not given
so, there it is. i didnt get very far into it.
 
  • #4
Doc Al
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Attack the problem step by step. Start, as always, by identifying all the forces acting on each of the masses. Draw a picture with those forces labeled. (All we care about are the horizontal forces.)

Once you've done that, there are several ways to proceed. Here's one way. Treat each mass separately and apply Newton's 2nd law. You'll get an equation for each mass. Combine them and you can solve for the tension in the cord and the fricitional force. Give it a shot.
 
  • #5
undertow2005
i did give it a shot, please i'm totally lost here, my Final that i must Ace is tomorrow. please work this problem out for me, somebody
 
  • #6
arildno
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What forces are the front block subject to?
Clearly, the applied force F, a tensile force T, and a frictional force f.
Obviously, both the tensile force and the frictional force works in opposite direction than F, so we have for particle 2:
[tex]F-T-f=ma[/tex]
For the rear block, the tensile force has the same magnitude T, but works in the opposite direction.
Hence, we have:
[tex]T-f=ma[/tex]
You have only 2 unknowns here, T and f, and 2 equations..
 
  • #7
undertow2005
from the equations i know how get get T and f. i have another question, how do you know T is opposite the force. i acutally thought T was going the same direction as the force for the box in the front. or is the tension always opposite the force. also, for the box in the rear, why did you leave out F out of the equation and not write F-T-f=m*a. i dont understand because the force is also acting on this box.
 
  • #8
Doc Al
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undertow2005 said:
i have another question, how do you know T is opposite the force. i acutally thought T was going the same direction as the force for the box in the front. or is the tension always opposite the force.
The tension in the cord exerts a pull (T) at both ends of the cord. According to your diagram, the cord pulls m1 to the right and m2 to the left.
also, for the box in the rear, why did you leave out F out of the equation and not write F-T-f=m*a. i dont understand because the force is also acting on this box.
The force F is only applied to mass m2; it does not touch m1. It will, of course, affect the tension in the connecting cord--but that's a separate force.
 

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