1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Force Problem with tension, 2 blocks, and a rope that has mass.

  1. Sep 23, 2010 #1
    1. The problem statement, all variables and given/known data

    Two blocks m1= 9.4 kg and m2 = 4.9 kg are connected by a homogeneous rope that has a mass of mr = 1.42 kg. A constant vertical force, F = 184.9 N, is applied to the upper block.

    What is the magnitude of the acceleration of the system?

    What is the magnitude of the tension in the rope at the bottom end of the rope?

    What is the magnitude of the tension in the rope at the top end of the rope?

    2. Relevant equations

    Newton's Second Law = [tex]\sum F[/tex] = ma


    3. The attempt at a solution

    For the first part, I took the objects to be part of one system, used F=ma with 184.9=F, combined the 3 masses, and solved for a. I then subtracted 9.8 m/s^2 to get 1.96 m/s^2.

    Tension Parts: I wasn't really sure how to do this from the get go, so I started trying some things. I'm also still not 100% sure about the relationships going on here. For the mag. of the tension of the rope at the bottom end of the rope, I took (4.9kg)(11.76 m/s^2) + (4.9kg)(9.8 m/s^2). I also tried it adding the mass of the rope on to the second block, so (5.62kg)(9.8m/s^2) + (9.4kg)(11.76 m/s^2)- I then wasn't sure if I needed to take into account block 1.

    Because I couldn't figure out the bottom tension, I didn't attempt to do the tension of the top of the rope.

    Any help would be appreciated. Thank you
     
  2. jcsd
  3. Sep 23, 2010 #2

    Doc Al

    User Avatar

    Staff: Mentor

    You managed to get the right answer, but a better approach would be:
    ΣF = 184.9 - Mg
    Set ΣF = Ma and solve for a. (M is the total mass.)

    Hint: To find the tension at the bottom of the rope, analyze the forces on the bottom block. The tension is one of the forces acting on that block. Apply ΣF = ma.
     
  4. Sep 23, 2010 #3
    Would the second force be weight?, w=mg (w=(4.9kg)(9.8m/s^2)
    and then possibly

    T-48.02=(mass of M1 + Mrope)(11.76 m/s^2) ?
     
  5. Sep 23, 2010 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Yes. There are two forces acting on the bottom block: the rope tension pulling up and the weight pulling down.

    The left side of the equation is fine, but the right side has two problems:
    (1) Since you're analyzing forces on the bottom block, the mass needed is just the mass of the bottom block.
    (2) The acceleration of the system is not 11.76 m/s^2. You found the acceleration in post #1. Use it.
     
  6. Sep 23, 2010 #5
    I found the tension of the bottom of the rope to be 57.6 N, and I'm working on the top part.
     
  7. Sep 23, 2010 #6

    Doc Al

    User Avatar

    Staff: Mentor

    Good.
     
  8. Sep 23, 2010 #7
    Do I assume the mass of the rope to be included with the mass of the first block? to give the force going up as (9.4kg + 1.42kg)(1.95 m/s^2) = 21.1 N
     
  9. Sep 23, 2010 #8
    What kind of martial arts do you train in/interested in?
     
  10. Sep 23, 2010 #9

    Doc Al

    User Avatar

    Staff: Mentor

    Not if you want to find the tension at the top of the rope. That end of the rope must be at the edge of your system. You can either: analyze forces on the top block or analyze forces on the 'rope + bottom block'.
     
  11. Sep 23, 2010 #10
    So, block 1 has a force going up which is 184.9 N. The force going down is weight. Is weight here just the weight of block 1 or the rest of the system of the rope and block 2.

    If so, the weight of the rope and block 2 has a force of 61.936 N. And the block going up has a force of 184.9 N. So added together the magnitude of the force is 246.84 N?
     
  12. Sep 23, 2010 #11
    Never-mind that is wrong too.
     
  13. Sep 23, 2010 #12
    If you are allowed to offer me any more hints, here is what I've come to until I'm stuck.

    In analyzing the top block, the force going upwards is 18.33 N (9.4kg)(1.95m/s^2). There is a force going downwards, the weight, which I'm not sure about. Is the weight needed for only block 1 or the weight of the rope and block 2 added together. In either case, if I'm doing the calculations correctly for either block 1 or block 2 + rope, I am getting the wrong answer.
     
  14. Sep 23, 2010 #13

    Doc Al

    User Avatar

    Staff: Mentor

    Don't call that a force, that's just mass X acceleration, which Newton's 2nd law tells us equals the net force on the object.
    The weight acting on the top block is just the weight of the top block. Note: There are 3 forces acting on the top block. What are they? Set their sum equal to 'ma'.
     
  15. Sep 23, 2010 #14
    Sorry, the computer says I finally got it right. I don't know why I kept getting caught up in thinking about the other objects instead of just block 1. You told me that I had calculated the net force when I multiplied block 1 (9.4kg)(1.95m/s^2) = 18.33 N, which is = ma. The weight of block 1 was (9.4kg)(9.8) = 92.12 N.

    T + 92.12 N = 18.33 N
    T = -73.79 N
    Magnitude = 73.79 N

    Thank you for you help.
     
  16. Sep 23, 2010 #15

    Doc Al

    User Avatar

    Staff: Mentor

    Sorry, but this solution is not correct. (Even though your answer is close.)

    Try this:

    ΣF = ma
    F - T - mg = ma (where F here is the applied force).
     
  17. Sep 23, 2010 #16
    sorry i have a similar problem to this; on the last part, are both of the "m"s the summation of all of the masses? so F - T - (summation masses x g) = (summation masses x a)?
     
  18. Sep 23, 2010 #17
    nevermind I got it, thanks for the help
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Force Problem with tension, 2 blocks, and a rope that has mass.
Loading...