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2 Blocks, 1 Cord [Acceleration, Tension]

  1. Oct 6, 2007 #1
    1. The problem statement, all variables and given/known data

    Two blocks connected by a cord passing over a small, frictionless pulley rest on frictionless planes:

    Which way will the system move when the blocks are released from rest? Right or left?
    What is the acceleration of the blocks?
    What is the tension in the cord?

    2. Relevant equations


    3. The attempt at a solution

    First off all, it should move to the left, as there is more weight there.

    We know that there is a force of gravity acting on each of the blocks straight downwards.

    And the pulling back of the block 100kg is not much, thus allowing it to slide to the left.
    For the 50kg block, the tension is greater than the normal force and the gravity force which allows it to move.

    The acceleration of the system is the total sum of the forces divided by the sum of the masses.

    The tension of the cord is the difference between the forces at the end of each.

    If my description is correct, how do I go into the mathematical part.?
  2. jcsd
  3. Oct 6, 2007 #2


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  4. Oct 6, 2007 #3

    Doc Al

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    What matters is the component of gravity down the incline.

    Tension and normal force are perpendicular to each other. What allows it to accelerate is a nonzero net force.

    True, if you do it right. But I advise against taking such shortcuts at first.

    What makes you think that?

    Rather than trying to do it in your head, attack it systematically:

    Examine the forces parallel to the incline acting on each mass. Apply Newton's 2nd law to each.
  5. Oct 6, 2007 #4
    ok I got the acceleration, with the tip you provided snoop. :D,

    now I will try tension. :)
  6. Oct 6, 2007 #5


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    You mean the shortcut? But I agree with Doc Al... I advise doing this from the basics...
  7. Oct 6, 2007 #6

    Actually that was the same equation I found in my textbook. I guess I should do it from the basics, so I can understand this more, than just plug in.
  8. Oct 6, 2007 #7


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    Yeah, I agree... the reason I brought that formula up is because we can get much more complex variations on the atwood machine... and in those cases it might not be practical to go through it step by step...

    But definitely in this problem, and while you're still learning about the atwood machine, you should do the problem from the basics.

    That thread is still very helpful. Dick gives good suggestions on how to solve the problem from the basics.
  9. Oct 6, 2007 #8
    I still do not seem to understand how to get the x and y components here. :(

    Sorry if you find this question annoying.

    I tried shifting the x and y graph so that the tension of the 100kg block in on the x axis.
  10. Oct 6, 2007 #9


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    I recommend reviewing this website if you're having trouble with the incline stuff:

    http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Dynamics/InclinePlanePhys.html [Broken]

    then try to apply that stuff here...
    Last edited by a moderator: May 3, 2017
  11. Oct 6, 2007 #10
    ok I was able to determine the acceleration of both objects.

    The one that has 100 kg has acceleration of 2.94 and the 50kg has 5.87.

    This must mean that the 50kg objects is going down and that the 100kg object is going up.

    So the blocks should slide to the right.

    Is this correct thus far?

    PS, I know realized that the values may be incorrect, I will keep on analyzing, but is the direction of sliding correct?
    Last edited: Oct 6, 2007
  12. Oct 6, 2007 #11

    Doc Al

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    The blocks are attached by the cord. How can the magnitude of their accelerations differ?

    How did you arrive at this?

    Did you analyze the forces acting on each mass? Apply Newton's 2nd law?
  13. Oct 6, 2007 #12
    Yes, I realized that it was wrong, since like you mentioned they are attached by a cord and should have the same acceleration magnitude wise.

    I got this by drawing them into different figures, and got the x and y component.

    I found that the x component of the 100kg object is 100(9.8) sin 30 and y component it was 100(9.8)cos30.

    For the 50kg object, the x of 100(9.8)sin 53.1 and y of 100(9.8)cos 53.1.

    This followed as well as I could understand the explanations listed in the similar scenario by Snoopy of getting the sum of Fx which gave me 293.69 and after applying the F=ma...I got different accelerations. This should not be the case.

    The forces acting on each mass is as follows, object 100kg has a gravity and normal force equal in size, and a tension to the right. object 50kg has gravity and normal force equal in size, and a tension to the left.
  14. Oct 6, 2007 #13

    Doc Al

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    The normal force doesn't equal gravity (which is mg) but it equals the y-component of gravity ([itex]mg\cos\theta[/itex]).

    Analyze each mass separately. (You'll connect them later by what you say about how their accelerations must be related.)

    We know that the masses only move in the x-direction, and that the y-direction forces balance. So, what forces in the x-direction act on m_1? On m_2?

    You can figure out the direction that they will accelerate by comparing the x-component of gravity on each mass. Which ever one pulls more, that's the direction of the acceleration.

    But you don't have to know the acceleration--or its direction--it will come out of the equations. If you want, just guess the direction and solve for the acceleration. You'll know by the sign of your answer if you picked the correct direction. (A negative sign means it goes opposite to what you thought.)
  15. Oct 6, 2007 #14
    Ok, I redrew it and analyzed the object of 100kg.

    I determined that it looks something like this.


    If this is correct, then the normal force should be 848.71 right?

    "So, what forces in the x-direction act on m_1?"

    The forces on x direction would be 490.

    I just did another one for the 50kg object, I found that the x component is 391.85 and the y is 294.21.

    "You can figure out the direction that they will accelerate by comparing the x-component of gravity on each mass. Which ever one pulls more, that's the direction of the acceleration."

    The one pulling more is the 100kg object. So it would slide to the left. (This is of course, if I determined the components correctly)
    Last edited: Oct 6, 2007
  16. Oct 6, 2007 #15

    Doc Al

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    It looks like you've got it. The x-component of gravity for each mass is [itex]mg\sin\theta[/itex].

    So the system accelerates to the left. Call the magnitude of the acceleration "a" and the tension "T".

    Let's deal with m_1 (100 kg) first: You have gravity acting down the incline, so that force should be called [itex]-m_1g\sin\theta_1[/itex]. (The minus sign indicates "to the left and down"; a plus sign means "to the right and up".) The tension is +T. So the total force is: T - [itex]m_1g\sin\theta_1[/itex]. Since the acceleration is down, we'd better call it -a.

    Combining all of this into Newton's 2nd law, we get:
    [tex]T - m_1g\sin\theta_1 = m_1(-a) = -m_1a[/tex]

    Now do the same kind of analysis for m_2.

    You'll end up with two equations. Solve them together and you'll get the acceleration and the tension.
  17. Oct 6, 2007 #16

    for m_2

    we have gravity acting down on the incline, so the force will also be called -m_2gsin(angle). I would say the tension is also +T (don't the tension have to be the same as they are the same cord). So far it is T - m_2gsin(angle). But now the acceleration is up, so it would be +a.

    When put together it would be.

    T - m_2 g sin (angle) = m_2 (a) = m_2.

    Solving for acceleration:

    T-m_1gsin(angle) = -m_1a
    T - m_2 g sin (angle) = m_2a.

    Ok, after pluggin in and solving for a and t

    I got that a = .654 and that t = 424.57

    hopefully this seems right. :)

    well I know that acceleration is right, because, I took the shortcut earlier and got the same answer, but now I did it with the basics and it feels great achieving this. Now I hope my T is right. :D
    Last edited: Oct 6, 2007
  18. Oct 6, 2007 #17

    Doc Al

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    Very good! Once you get the hang of the "from the basics" method, you'll be able to solve all sorts of problems without having to memorize any special results. It's the way to go.

    Here's your assignment. Study the relevant examples on this page: Standard Newton's Laws Problems
  19. Oct 6, 2007 #18
    Doc Al, thank you for taking time to teach me the basics.
    I will study the examples on the page you provided, so I can get better at this and one day master physics.:smile:
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