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Friction problem of rope

  1. Apr 16, 2008 #1
    1. The problem statement, all variables and given/known data
    A student pulls on a rope attached to a box of books and moves the box down the hall. The student pulls with a force of 185N at an angle of 25.0 degrees above the horizontal. The box has a mass of 35.0kg and a mu between the box and the floor is 0.27. Find the acceleration of the box.

    2. Equations:
    sum of Ftx=Ftx+Ff=ma
    sum of Fy= Fn+Fpy-Fg=0
    Ftx=Ft cos 25 degrees
    Fty= Fy sin 25 degrees

    3. The attempt at a solution

    In class we drew a picture and wrote out many, many equations that we are supposed to use to solve the problem. Each equation is supposed to help us solve another equation. I am very confused. Does anyone know another way to solve this problem or explain it differently. I have written down 168N-71.5/35kg and got 2.75 which from what my teacher says is the correct answer. I can not remember how I did it and don't know if what I did is correct.
  2. jcsd
  3. Apr 16, 2008 #2
    First step to any force problem is to draw a free body diagram. Have you gone this far?
  4. Apr 16, 2008 #3
    What follows is a rather lengthy monologue on these types of problems. I hadn't originally planned to type so much, but it just turned out that way. If a lot of what I say you already know, that's great, but the following thoughts have helped clarify mechanics in my mind as I learn it also.


    This problem does have several steps, and there is no way around that, but we can try to help you understand those steps by going through it step by step again. I think that after being exposed to it a few times, you will understand what is going on, and when you see another similar problem, will be able to work out the steps logically by yourself. When you become familiar with it, one step really leads to the other.

    In a typical mechanics problem like this, when you are asked to find acceleration, that is the same thing as telling you to find the net force on the object, because then you can find the acceleration with F = ma.

    What is the net force? Well, what forces are acting on the box? In what directions? Draw a free body diagram. It turns out that there seems to be four forces on the box. One is from the gravity pulling the box toward the center of the Earth--it is really to the center of the Earth, but from our perspective, that is down. One is the normal force--the support the ground gives to the box so that it does not move vertically. One is the force applied by the student, which we are told is 185 N in a direction 25 degrees above the horizontal. Let's draw this pointing to the right. Finally, there is the friction force pointing horizontally to the left. Note that if we had drawn the student's force pointing to the left, the friction force would be to the right, since it always opposes the direction of motion.

    A simple thing to think about at this point is why the friction force is pointing horizontally. Since the student's force is pointing 25 degrees up to the right, why is the friction force not pointing 25 degrees down to the left? Well, for one thing, you probably already know that friction is, in rough terms, caused by two things rubbing against one another. If the student is dragging the crate, it would be sliding across the ground horizontally to the right, so the friction force must be opposite: horizontally to the left.

    Here's a question that follows: how come the student's force, which is 25 degrees upwards, cause the crate to move horizontally and not in line with the force? You can see that it is the weight of the box that is too much for the upward part of the force to overcome. To be more quantitative, you need to break the student's force into components, so you can analyze the vertical component of it versus the box's weight and the horizontal component versus friction.

    Something you may ask is, why are the components vertical and horizontal? Can they be rotated, say the horizontal axis be 25 degrees up and the vertical axis perpendicular to that? Well, yes it can, and the math works out, but it turns out that it is harder to do that because then the unknown friction force needs to be broken up into components and the weight and normal force both are broken into components, and you will end up with a lot more variables.

    It takes some experience and insight to see that using horizontal and vertical axes is the best for this problem, but that is the case. This is because, as I glossed over above, you only need to break one force, the student's into components, rather than break 3 forces (weight, normal, friction) into components.

    You may also ask, why do the axes need to be perpendicular to each other? It's because only perpendicular (e.g. horizontal and vertical) motions do not influence each other in each way. Thus, we can break any complex motion into two directions perpendicular to each other, analyze them separately, then put it back together if we need to. Is there a deep physical reason why this is true? Maybe. I don't know, but that is what we have found out about this world.

    Back to the problem. We know there are four forces. Now, we need to analyze them in the vertical and horizontal direction, in which case force values can add and subtract together (you could not just numerically add forces in different directions). And remember that this is good because the horizontal and vertical motions are completely separate from each other, so horizontal force values do not influence vertical force values, and the problem is simpler now (by that, I mean you have a direction in which to move forward, that is, by analyzing the horizontal and vertical forces and motion).

    What are the forces you do not know? Do you need to find what they are? How do you find what they are? Try to go on from there.
  5. Apr 16, 2008 #4
    Thank you for your explanation. It did help me to better understand the concepts. However, I am still confused as to how I got the number 71.5 in the equation, 168N-71.5/35 kg=a. I did something and did not write it down and can not remember what formula I used.
  6. Apr 16, 2008 #5
    That is the friction force, which is equal to [itex]\mu_k N[/itex]. What is N?
  7. Apr 16, 2008 #6
    N= 343-78.2
  8. Apr 16, 2008 #7
    Right. By experiment, we have found that the friction force in normal cases is proportional to N, where the constant of proportionality is mu, so that [itex]f = \mu_k N[/itex] (when the object is in motion, we use the coefficient of kinetic friction). You are told mu (in real life, you may find mu in a table for two materials). Thus, you can find the friction force. Now, do you need to use that friction force to answer the question (you already used it in your equation, but make sure you understand why).
  9. Apr 16, 2008 #8
    Thank you. Now I see how to get the answer.
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