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Frictional force, velocity, and acceleration

  1. Feb 20, 2012 #1
    1. The problem statement, all variables and given/known data
    A rope is tied to a log of mass m = 99.9 kg. When you pull on the rope with a horizontal force of magnitude F = 688 N, the log moves at constant velocity.

    a) Find f, the magnitude of the resistive (frictional) force acting on the log. You may assume the frictional force acting on the log is constant.

    b) Find the magnitude of the horizontal force F needed to pull the log with an acceleration of magnitude a = 5.42 m/s2. Assume the resistive force is the same as in part a.


    2. Relevant equations
    F = ma


    3. The attempt at a solution
    I'm actually stuck on the fact that the problem says that when we pull with a force of 688N, the log moves at constant velocity. Wouldn't that mean that there is (a) no friction, and (b) acceleration = 0? Obviously I'm wrong, but that's how I'm interpreting it, and I sort of have a mind block that won't let me get around it.

    Even if we solve for acceleration, F = ma

    688N / 99.9 kg = 6.8869 m/s2

    Would we have to find a frictional force that would make this acceleration 0 and the velocity constant?
     
  2. jcsd
  3. Feb 20, 2012 #2

    Redbelly98

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    Welcome to Physics Forums.
    Constant velocity does mean acceleration = 0, and hence the net force (∑F) is zero. You can't say that any particular force is zero, only that the vector sum of all forces acting on the log is zero.

    Have you drawn a force (free body) diagram for yourself?

     
  4. Feb 20, 2012 #3
    Thank you for the welcome and response.

    I did draw a free body diagram, but I might be doing it wrong. I just have a mass of 99.9kg that is being pulled on by a force of 688N to the right (F1). As you said, the net force should be equal to 0, so what I have set up now is another force going to the left to show that it is a frictional force (Ff). Now I have this set up:

    F1+Ff = m*a
    688N+Ff = (99.9kg)*0 (because constant velocity means 0 acceleration)
    (688N+Ff)/(99.9kg) = 0
    Ff = -688

    That doesn't seem quite right because if I am pulling on the log at 688N, and the frictional force is -688, then the log is receiving complete resistance and isn't moving. There's something wrong with my calculations, and I can't see it.
     
  5. Feb 20, 2012 #4

    Redbelly98

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    It's good that you are thinking about the implications of your results.

    The fact that the two forces add to zero means that the acceleration, and not necessarily the velocity, is zero. What happens in reality is that a force greater than 688 N is applied at first to get the log moving, and then the applied force will be 688 N after it starts moving, in order to maintain a constant velocity.

    So your calculation of the friction force (688 N in the opposite direction as the applied force, i.e. -688 N) is correct. :smile:
     
  6. Feb 20, 2012 #5
    Thank you very much, Redbelly98! This part one of a two-part question, and I needed that part solved correctly before I could answer the other. I got them both correct, so I'm glad.

    I do admit that I have two more questions which I'm stuck on, and I'm hoping it's alright to post those questions here instead of making a new thread. Please let me know if I should make a new one.

    Here they are:

    1. The problem statement, all variables and given/known data
    Mass m sits on a frictionless, horizontal table. Assume:
    - All vertical forces acting on m (such as gravity and the normal force) sum to zero.
    - All applied forces act parallel to the table
    When a force of magnitude F is applied to mass m, it accelerate at magnitude a = 4.49 m/s2.

    NOTE: The axes lie along the table.

    Suppose force F1 has magnitude 2F and points at 45 degrees to the x-axis. If forces F1 and F2, with a magnitude F, act on mass m at the same time, what will the magnitude of its acceleration be now?


    2. Relevant equations
    Components


    3. The attempt at a solution
    I attempted to solve this by drawing a diagram, and F1 is pointing out at 45 degrees relative to the x-axis. I found its components, with the x-direction vector being 8.98cos(45) = 6.3498. F2 is already in that direction as well, so I just added the two vectors to get a sum of 10.8398. When I entered this answer as 10.84 into Webassign (online homework - I don't know who is familiar with it), it said that the answer is incorrect. Is it that I didn't put enough digits, or did I solve the problem incorrectly?



    1. The problem statement, all variables and given/known data
    A red box and a blue box sit on a horizontal, frictionless surface. When horizontal force F is applied to the red box, its acceleration has magnitude a = 5.09 m/s2.

    a) If force F is applied to the blue box, its acceleration has magnitude a = 1.24 m/s2. Find mred/mblue, the ratio of the mass of the red box to the blue box.

    b) Now, the two boxes are glued together. If horizontal force F is now applied to the combination, find a-combo, the magnitude of the acceleration of the combination of boxes.


    2. Relevant equations
    F = ma


    3. The attempt at a solution
    I already correctly answered question a, which came to be a ratio of .2436. The second part, however, always confuses me because there has to be some sort of manipulation. Intuitively, when we put together the two boxes, we are combining their masses, making the total acceleration even smaller than what they were individually. This is true because we are applying the same amount of force to both of the objects. To get the numerical answer for the combined acceleration, I have no idea. I already know that adding the two accelerations together (two vectors in the same direction) and then multiplying it by the ratio (6.33 m/s^2 * .2436) gives me an incorrect answer, so that is probably a poor approach. I'm just stuck, like the rest of the problems I've had. lol it's quite frustrating.
     
    Last edited: Feb 20, 2012
  7. Feb 20, 2012 #6

    Redbelly98

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    Hi again,

    Please make new threads for each new question.
     
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