Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Net Force / Friction on Blocks

  1. Sep 28, 2005 #1
    Consider two blocks, A and B on three surfaces. On each surface, block A is smaller than block B, and always positioned to the left (they are touching, with block B on the right). We have three situations:

    1. Block A is pushed to the right on a smooth surface.
    2. Block A is pushed to the right on a rough surface. The blocks are at rest, and force F is smaller than the maximum static friction force on A.
    3. Block B is pushed to the left. The blocks move at constant speed.

    The force F is assumed to be the same for each. When moving the blocks will encounter air resistance.

    My problem here is deciding two things: In order from highest to lowest, I need to order them depending on the net force on block A. Also, I need to order them in the same way depending on the total friction exerted on A and B.

    After drawing the free-body diagrams, it looks like there is the most net force on A in situation 1 (due to only wind resistance), second most in situation 3 (due to wind resistance and a small amount of kinetic friction), and the least in situation 2 (due to the high amount of static friction). Does this look alright (as far as my reasoning goes) or am I not including something?

    For the second problem, I am saying 2>3>1 because 2 doesn't move at all due to the high static friction, 3 has both wind resistance and kinetic friction, and 1 has only wind resistance. Again, is this a reasonable answer?

    I would appreciate your help in this (although please don't give away the answer to me) :smile:

    Attached Files:

    Last edited: Sep 28, 2005
  2. jcsd
  3. Sep 28, 2005 #2


    User Avatar
    Homework Helper

    For part 1), if an object is static, or moves with constant velocity, i.e. no acceleration, then the net force on the body is zero.

    For part 2), that ordering and reasoning are correct/good.
  4. Sep 28, 2005 #3
    Hmm... Looking at it again it shows that my answer for question 2 is incorrect. I guess I am taking air resistance into account wrong somewhere...

    Edit: On diagram 1, I have only wind resistance (which I am taking into account as friction) working only on block B. In diagram 2, I have only static friction, which is greater than the force F exerted on the side of the block. In diagram 3, I have wind resistance working on block A and part of B, and also some kinetic friction from the surface. I am having trouble ranking these however, without and values given. Does anyone have a suggestion on how to do this (apparently my reasoning from above was incorrect :confused: )?

    Thank you.
    Last edited: Sep 28, 2005
  5. Sep 28, 2005 #4
    Actually, now that I think of it, since the blocks in situation 3 are moving at a constant speed -- does this mean that the friction forces cancel out?
  6. Sep 29, 2005 #5


    User Avatar
    Homework Helper

    Hmmm ...
    I don't understand why it told you your answer was wrong.

    In Fig 1, there may be some wind resistance, but the blocks are moving on a smooth (AKA frictionless) surface, therefore there is no friction and <1> is last in the order - as you had it.

    In Fig 2, there is no wind resistance and no movement, so the static friction is simply equal to the applied force, so Fr,2 = F. (Fr,2 is the total friction on blocks in Fig 2)

    In Fig 3, the block is moving at a constant speed, there is no acceleration, so the net force on the blocks is zero, so the forces must balance. The only forces are wind resistance, R, the applied force, F, and kinectic friction, Fr,3. (Fr,3 is the total friction on blocks in Fig 3)

    The balance of forces is,

    F = R + Fr,3

    substituting for F = Fr,2

    Fr,2 = R + Fr,3
    => Fr,2 > Fr,3

    So, the ordering of (total) frictional force on the blocks is

    <2>, <3>, <1>
  7. Sep 29, 2005 #6
    Yeah that's what I was thinking as well, which is why I cannot see the reason for which it is incorrect.

    Thanks for the help Fermat.
  8. Sep 29, 2005 #7
    Hmm... Maybe the fact that block A is smaller than block B?
  9. Sep 29, 2005 #8


    User Avatar
    Homework Helper

    The question asked for "the total friction exerted on A and B".

    I don't think that the diffrence in sizes makes a difference.
  10. Sep 29, 2005 #9
    Well, look at the first situation. Block A doesn't experience any friction but block B does (from the wind). By the way, in this case is wind considered a friction force?
  11. Sep 30, 2005 #10


    User Avatar
    Homework Helper

    OK, that may be my mistake - I think.
    Air resistance is a form of friction, but I usually think of it as being referred to as drag.
    I took the wind resistance here as being a resisting force rather than a frictional force.
    If we consider the wind as contributing to the total friction, then what do we have ?

    Part 2)

    Case 1: The blocks are on a frictionless surface and have a force F applied to them. Let R1 be the wind resistance.The net force on the blocks is F - R1, so there is an accelerating force of (F-R1) which will increase the velocity of the block combination. Now drag, or wind resistance, is proportional to the velocity (or, the square of the velocity) and so R1 will increase in value as the blocks increase in speed until terminal velocity is reached, which is when R1 is equal to F and then there is no longer any net force acting and the blocks will continue at constant speed.
    If the wind resistance is taken as friction, then the total friction on the blocks, in case 1, is
    Friction1 = R1 = F

    Case 2: The blocks are static, there is no wind resistance, only surface (static) friction, so the total friction on the blocks is the applied force,
    Friction2 = F

    Case 3: The block combination is moving with constant speed, so no net force. The wind resistance and surface (kinetic) friction balance with the applied force, F. Let R3 be the wind resistance and Fk be the surface (kinetic) friction. Then R3 + Fk = F
    i.e. total friction on the blocks is,
    Friction3 = F

    Now, according to this analysis, all cases have the same amount of total friction, therefore they cannot be ordered - which is a bit crazy !!

    I still can't see how the different block sizes make any difference. Drag will alter according to the profile being presented to the wind flow. But as far as I can see the profile of the block combination is the same in either direction.

    Can you tell me where this question came form ?
  12. Sep 30, 2005 #11
    It was a problem from my physics class, which I still cannot get. The TA told me that I have the correct reasoning (2>3>1), but there was a misprint in the problem and it was supposed to say that figure 3 is smooth also, thus 2>1=3. However, when I entered this into the system (it was an online homework), it told me once again that I was incorrect.

    It's just a bad problem.

    Thanks again for your help.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook