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Homework Help: Pulley Systems! HELP

  1. Jul 6, 2010 #1
    Pulley Systems! HELP!!

    1. The problem statement, all variables and given/known data
    The three blocks in the diagram are released from rest and accelerate at the rate of 1.5m/s^2. What is the magnitude of the frictional force on the block that slides horizontally?

    i dont know why my picture isnt attaching but basically the image is of a table and on it is 1 block(4kg), on the left dangling off the table with a pulley is another block (2kg), and the same on the right side except with a mass of 4kg



    2. Relevant equations
    Ff=MFn
    f=ma


    3. The attempt at a solution
    so...i am very confused, i believe that im trying to find the frictional force of the block on the table
    i thought that the force of gravity and normal force cancel...and so what's left is the tension from the left and right, as well as the frictional force from the right...
    the total force of the system is equal to the (total mass)(acceleration), which is 15N
    now what? what do i do?!
     
  2. jcsd
  3. Jul 6, 2010 #2
    Re: Pulley Systems! HELP!!

    The frictional force opposes the motion of the box. Which way is the block on the table accelerating?
     
  4. Jul 6, 2010 #3
    Re: Pulley Systems! HELP!!

    im guessing towards the right...with the block with the most mass?
     
  5. Jul 6, 2010 #4
    Re: Pulley Systems! HELP!!

    So the frictional force is to the left? correct? What else besides the frictional force is acting on the block(on the table) to the LEFT?
     
  6. Jul 6, 2010 #5
    Re: Pulley Systems! HELP!!

    the tension from the 2kg block dangling off the pulley to the left?
     
  7. Jul 6, 2010 #6
    Re: Pulley Systems! HELP!!

    Yes indeed. So those two forces are acting to the LEFT. One Force is known, the other is not, friction. What forces are acting on the block to the RIGHT?
     
  8. Jul 6, 2010 #7
    Re: Pulley Systems! HELP!!

    how is tension known?

    and from the left is the tension from the left block...the force of gravity and normal force cancel out...
     
  9. Jul 6, 2010 #8
    Re: Pulley Systems! HELP!!

    btw thanks for taking the time to help me out, much appreciated!:)
     
  10. Jul 6, 2010 #9
    Re: Pulley Systems! HELP!!

    The block on the left must accelerate up. So T beats out mg by ma for the left, yes? In other words, Tension acting up on the 2kg mass is larger than mg acting down on the 2kg mass because the 2kg must accelerate up if the 4kg block accelerates to the right.

    So T - 2g = 2a... you know g and a...

    You really are best off doing this using free body diagrams and keep a strict accounting of all forces and the net force. Have you used these before?
     
  11. Jul 6, 2010 #10
    Re: Pulley Systems! HELP!!

    oh wow
    i never knew there was a tension equation...
    i do have a free body diagram...and so if i have the tension(and they are equal on either side of the block, right?) and my only left unknown is friction...do i just say Ff+T=ma???
    and then if they are equal on either side...dont they cancel??
     
  12. Jul 6, 2010 #11
    Re: Pulley Systems! HELP!!

    There is not a tension equation. Tension is considered a force. Look at this FBD... and see if it is understandable how I got T - mg = ma or T = mg + ma
    fbd.JPG

    This is a FBD for the 2kg mass. There is a net force on it. It must be accelerating up. The net force is just ma
     
  13. Jul 6, 2010 #12
    Re: Pulley Systems! HELP!!

    ohhh ok i get itt
    so then the net force of that block is equal to ma, and the net force is equal to only the force of friction? because the tensions cancel each other out?
    but that cant be right...the answer is too easy that way...
     
  14. Jul 6, 2010 #13
    Re: Pulley Systems! HELP!!

    no...

    Can you draw a FBD for the 4 kg block on the right? and then give me the equation you used to find this T on the right? its diff from the T on the left otherwise the 4 kg block on the table would not accelerate to the right. The T on the right must be bigger than the T on the left.
     
  15. Jul 6, 2010 #14
    Re: Pulley Systems! HELP!!

    OMG I GOT IT!
    ok:
    Ft(from 1)=ma+fg(of 1)
    Ft(from 3)=fg(of 3)-ma

    Ff(of 2)=Ft3-Ft1-ma
    plllllzzz tell me im right! lol i hope you understood what i wrote
     
  16. Jul 6, 2010 #15
    Re: Pulley Systems! HELP!!

    Does it make sense that the tension on the block on the left is T = 2(10) + 2(1.5) ?
     
  17. Jul 6, 2010 #16
    Re: Pulley Systems! HELP!!

    yup! so it's right then?
    thank you SO much...could you just answer one more question for me, really quickly, i just need to clear up something
     
  18. Jul 6, 2010 #17
    Re: Pulley Systems! HELP!!

    What are 1 and 3? Is 1 the block on the left, the 2kg block and 3 is the block on the right, the 4 kg block? and I guess 2 is the block on the table?
     
  19. Jul 6, 2010 #18
    Re: Pulley Systems! HELP!!

    oh yeh sry but yeh
    1:the 2kg dangling off the left of table,
    2: the 4kg, main one on the table
    3: the 4kg, dangling off the right of the table
     
  20. Jul 6, 2010 #19
    Re: Pulley Systems! HELP!!

    So it looks like you got it. But please, please look at my FBD for the block on the table. You need to be able to do this accurately because some of these problems can be cumbersome if you dont. I have examined all the forces in the x and y directions and then from my FBD's I have written some equations. Tell me you get this first.
    fbd1.JPG
    Fn is the normal force.
     
  21. Jul 6, 2010 #20
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