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Magnitude of friction force between two boxes

  1. Oct 23, 2013 #1
    Hello! I'm excited to join this physics community and appreciate any help that the people posting in this thread will give me!

    The problem is this

    You are lowering two boxes, one on top of the other, down the ramp with length 4.75m and height 2.5m (The ramp has its highest point at the right. Even though this isn't relevant to the problem i wanted to include it to give the clearest picture of the situation considering i cannot post the figure) by pulling on a rope parallel to the surface of the ramp. Both boxes move together at a constant speed of 19.0cm/s . The coefficient of kinetic friction between the ramp and the lower box is 0.410, and the coefficient of static friction between the two boxes is 0.777.

    The mass of the upper box is 32kg and the mass of the lower box is 48kg

    Attempt at a solution:
    I was successful in finding the force needed to accomplish this task is 80.7N

    The trouble area is in the second part of the question where they ask "What is the magnitude of the friction force on the upper box?"

    So using trig i can see that the angle of elevation is around 27.759degrees

    Thus the normal force acting on the upper box would be the force of the weight of the box multiplied by the cosine of the angle of elevation giving me the perpendicular-to-the-plane component which = 32kg(9.8m/s^2)COS27.759

    The coefficient of static friction between the two boxes is .777 so the magnitude of friction acting on the upper box should be .777(32kg(9.8m/s^2)COS27.759) = 215.62N

    This answer is wrong and I have become incredibly frustrated as everywhere I look it seems i'm doing this correctly, so i must be missing something. Once again, any help is greatly appreciated.
     
  2. jcsd
  3. Oct 23, 2013 #2

    haruspex

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    Which box is the rope attached to?
    Remember that the coefficient of static friction only imposes a limit on the static frictional force. The actual force will only be whatever is required to balance the force tending to make the surfaces slide against each other.
     
  4. Oct 23, 2013 #3
    The string is attached to the bottom box, my mistake not including this initially.

    So the magnitude of the frictional force will be affected by the force of the rope on the system?
     
  5. Oct 24, 2013 #4

    haruspex

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    It depends. If you were to pull harder, causing the boxes to accelerate, but the top box still did not move relative to the lower box, then the frictional must have increased in the direction of acceleration, since no force on the top box has changed. (Note that if the frictional force is initially up slope then, at first, this means it will be reduced in magnitude.) If you pull hard enough the frictional force will hit its static limit and the top box will slide off the bottom box.
    So, consider the FBD of the top box. What are the forces on it and what is its acceleration?
     
  6. Oct 24, 2013 #5
    Ah ok then I'm guessing that would not apply in this scenario as the boxes are moving together at constant velocity and this the acceleration of the boxes is 0?

    When considering the FBD of the top box I include the following forces
    -friction force
    -gravitational force (two components are normal force and down slope force)
    -I'm uncertain how the force acting on the lower box pulling it up the slope will affect the upper box

    Since as stated in the problem both boxes move together at a constant velocity then the acceleration of the system and of the top box is 0?
     
  7. Oct 24, 2013 #6

    haruspex

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    The gravitational force acts vertically downward. The normal force from the lower box on the upper acts normal to the slope.
    The boxes are being pulled down the slope, I thought.
    The upper box is blithely unaware of the rope. It feels the effect via the frictional force where it contacts the lower box. There are only three forces to consider on the top box.
    Yes.
     
  8. Oct 24, 2013 #7
    I understand this i was simply saying that the normal force is derived by splitting the gravitational force into two components: one perpendicular to the slope (which is equal to the normal force) and one parallel to the slope which i called the down-slope force (Granted this may be the incorrect name for the parallel component force).

    I thought this was the case and I feel this is what is reflected in my calculations above yet my answer is still incorrect.

    YF-05-53.jpg

    Here is the image of the problem. This is driving me insane i cannot see what i'm missing here.
     
  9. Oct 24, 2013 #8

    haruspex

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    Right, and since it is not accelerating, what is that downslope component of gravity equal to for the top box?
    Ok, I understand the set up now. The pull on the rope is just sufficient to stop the boxes accelerating down the slope. But whether the pull is downslope or upslope makes no difference to the question about frictional force on the upper box.
     
  10. Oct 25, 2013 #9
    The down slope component of the top box should be the :

    (Mass of the box)(gravity)Sin(angle of elevation)

    32kg(9.8m/s^2)SIN(27.759)= 146N
     
  11. Oct 25, 2013 #10

    haruspex

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    Yes, that's its value, but in terms of combinations of other forces what is it equal to?
     
  12. Oct 25, 2013 #11
    Hey so 146 N Is the answer which makes sense considering that this force is required to keep the box from moving, but why Is this not the value of my original calculations? Shouldn't the answer be the same whether evaluating the force of friction derived from the normal force and from the coefficient of friction or from finding it by seeing what force is needed to keep the box from moving?
     
  13. Oct 25, 2013 #12

    haruspex

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    I thought I explained this. The coefficient of static friction only tells you the maximum ratio between the frictional force and the normal force. If the actual frictional force is less than ##\mu_k N## then sliding does not occur. You were asked for the actual force, not the maximum force.
     
  14. Oct 25, 2013 #13
    Ahhhhhh it all make sense now. Sorry I was so dense previously. Thank you so much for your help this is an important concept I would have missed otherwise!
     
  15. Oct 25, 2013 #14

    haruspex

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    It's a common misunderstanding.
     
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