1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Friction Force Homework Problem

  1. Jul 14, 2006 #1
    Hello, i have been asking alot of question lately, i had 60 questions to do for hw, and i'm kinda confused on some of them...

    A 0.5kg block is placed on top of a 1.0kg block, the coefficient of static friction between the two blocks is 0.35. The cofficient of kinetic friction between the lower block and the lever table is 0.20. What is the maxinimum horizontal force that can be applied to the lower block without the upper block slipping?

    Some Thoughts:
    First i found out the kinetic frictional force,between the lower block and the table. (i don't know if you should add the two masses together, since block 2 is on top of block 1, so does it change block 1's normal force? anyways i didn't add 0.5kg.)

    Block 1: mg(us) = (0.5kg)(9.8N/kg)(0.35) = 1.715N
    Block 2: mg(uk) = (1.0kg)(9.8N/kg)(0.20) = 1.96N

    when Force applied = Frictional, no accleration.
    So, F applied must first overcome block 2's frictional force, then you have a net force, but that force must not be greater than 1.715N

    So i added the number forces: 1.96N + 1.715N = 3.675N. The force applied must be less than this... but i was wrong, our book says the correct answer is 8.1N... i wonder how they got that.

    Thanks.
     

    Attached Files:

    Last edited: Jul 14, 2006
  2. jcsd
  3. Jul 14, 2006 #2

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    If you apply enough force to get the blocks moving at some small speed (no acceleration), what is the horizontal force that the lower block exerts on the upper block? What is it if you apply a greater force?

    Draw a free body diagram showing the horizontal forces acting on the upper and lower blocks. Set the force on the lower block to be just a bit greater than the force of static friction between the upper and lower block. What happens?

    Also, you have to add the weight of the two blocks for the normal force on the table surface (to determine the force of kinetic friction between the table and the lower block).

    AM
     
  4. Jul 14, 2006 #3
    Sorry i forget to sent the diagarm along with it.

    what is the horizontal force that the lower block exerts on the upper block?

    What do you mean by that? they are perpendicular, shouldn't it have no effect on each other?
     
  5. Jul 14, 2006 #4

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    I meant: when you apply a horizontal force to the lower block, what is the horizontal force that the lower block exerts on the upper block?

    AM
     
  6. Jul 14, 2006 #5
    umm, the same? as the lower block?
     
  7. Jul 14, 2006 #6

    Doc Al

    User Avatar

    Staff: Mentor

    I will ask the question differently. First question: What is the means by which the bottom block is able to exert a horizontal force on the top block? (What kind of force is it?)
     
  8. Jul 14, 2006 #7

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    No. Think of it this way: Assume that the force applied to the lower block is greater than the force of kinetic friction. What happens to the lower block? What is the horizontal force on the upper block from the lower block if they both move together? (hint: Newton's second law of motion applied to the upper block).

    AM
     
  9. Jul 15, 2006 #8
    oh wait, i get it now, well i got the right answer but i don't know why i did that, could someone explain to me why i did that??

    OK: First the Horizontal Force required to start the movenment, that's
    (1.0kg+0.5kg)*(9.8)*(0.20) = 2.94N

    Then in order for the block to fall off, it must over come the static friction between the two blocks, which is (1.0 + 0.5)*(9.8)*(0.35) = 5.145N

    Than you add the two together that's should be the maxinimum force allowed without the blocks slipping...BUT what i don't understand is that why do i have to add the mass of the lower block AND the upper block together to get the force of static friction? isn't it only suppose to be (0.5)*(9.8)*(0.35), without the lower block? i'm confused on this part.

    PS: yea i understand the horizontal part now, but not static...
    THANKS
     
  10. Jul 15, 2006 #9
    You want to start the system of the two blocks moving. The normal force for the two blocks is proportional to the sum of their masses. The acceleration of the two blocks also needs to consider both their masses. When you consider the maximum force before the top block moves relative to the bottom block, then you consider the masses seperately.
     
  11. Jul 15, 2006 #10

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    The static friction force is the MAXIMUM force that the lower block can exert on the upper block (horizontally). So let the force on the lower block be such that the two blocks stay together:

    [tex]F = (m+M)a + \mu_k(m+M)g[/tex]

    (1)[tex]a = \frac{F}{m+M} - \mu_kg[/tex]

    Since the static friction force has to supply the accelerating force on the upper mass: [itex]ma < \mu_s mg[/tex]

    So putting that condition into (1):

    [tex]\mu_smg > ma = \frac{Fm}{m+M} - \mu_kmg[/tex]

    So:

    [tex]F < (M+m)(\mu_sg + \mu_kg)[/tex]

    [tex]F < 1.5*(.35 + .2)*9.8 = 8.1[/tex]N.

    AM
     
    Last edited: Jul 15, 2006
  12. Jul 16, 2006 #11
    i see i see, thank you, the way you did that seems simple. Yet complicated, with all the variables, but i did understood that,thanks.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?