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Force with kinetic friction

  1. Mar 3, 2007 #1
    1. The problem statement, all variables and given/known data
    The figure shows 3 crates being pushed over a concrete floor by a horizontal force f of magnitude 440N. The masses of the crates are m1=30 kg, m2=10kg, and m3=20kg. The coefficient of kinetic friction between the floor and each of the crates is 0.7. what is the magnitude F32 of the force on crate 3 from crate 2? (a picture is attached)


    2. Relevant equations
    F=ma
    F_x=ma_x
    F-f_k=(m1+m2+m3)a_x
    f_k=μ(m1 + m2 + m3)g




    3. The attempt at a solution
    I solved for acceleration. my formula for acceleration is attached. hehehe. (oh, by the by, that formula is supposed to have "g" multiplied to the mu in the numerator). I got 0.473 m/s^2 for my acceleration. I read somewhere that to get F32, I should multiply the acceleration with m3. Somehow, that doesn't seem right to me. I ended up drawing a picture with space in between each crate. And F32 I conjured up to equal f_k (the kinetic force of friction) + F12 (force on crate 1 from crate 2) + m2*a_x. I then got a value of 402.12 N.

    F32 = f_k + F12 + m2*a_x = 402.12 N

    I am really lost. Please help me :frown:
     

    Attached Files:

  2. jcsd
  3. Mar 3, 2007 #2

    Doc Al

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    Staff: Mentor

    That's fine.

    To find the forces that the crates exert on each other, apply Newton's 2nd law to each crate separately. You only need to analyze one of the crates to find the needed answer, but I won't tell you which one. :wink: Hint: Consider Newton's 3rd law as well.
     
  4. Mar 3, 2007 #3
    ooh, thanks :D

    i'm analyzing crate 3. so F32 = m3*a_x.
    Therefore, F32 = 20.0 kg * 0.473 m/s^2 = 9.46 N?
     
  5. Mar 3, 2007 #4
    if i wanted to consider the middle crate, sort of like i was originally trying, would:

    F32 = f_k + F12 + m2*a_x

    ---would that be right?

    :D
     
  6. Mar 3, 2007 #5

    Doc Al

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    Staff: Mentor

    To analyze crate 3 (good choice) you must consider all the forces acting on it.


    That depends on how you defined your sign convention and your variables. For example, does F32 represent the magnitude of the force of crate 3 on crate 2? What direction does it act? Is that direction positive or negative?
     
  7. Mar 3, 2007 #6
    okies, i did not see that the question was actually asking for magnitude. woops.

    let's focus on crate 3. so i'd have to do the x component/y component situation. but, wouldn't the y component be zero? normal force minus m3*g?

    so then for the x component:
    F_x = F32-F23-f_k
    with F23 being magnitude of force on crate 2 from crate 3.

    so magnitude equal F_x? :biggrin:
     
  8. Mar 3, 2007 #7
    hmm...worked on it some more, and i'm getting:

    F32 = -m3*a_x-f_k with f_k=(mu)_k * m3*g
    = -(20kg*0.473m/s^2)-(0.700*20kg*9.8m/s^2)
    = -147 N

    therefore, magnitude is 147 N

    -possibly :eek:
     
  9. Mar 4, 2007 #8

    Doc Al

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    Staff: Mentor

    Right--the y-component of the net force will be zero.

    Since you're considering crate 3, only consider forces on crate 3. F23 would be irrelevant.


    Much better, but I don't know where those minus signs come from.

    Horizontal forces on crate 3, which moves to the right:
    (1) the contact force from crate 2: F32 acting to the right
    (2) the kinetic friction, acting to the left
    So, using + to mean "right", the net force is: F32 - f_k
    Applying Newton's 2nd law gives you:
    F32 - f_k = m3*a

    So: F32 = m3*a + f_k = m3*a + (mu)_k*m3*g
     
  10. Mar 4, 2007 #9
    hey cool!

    erm...those negative signs came from the way i drew my free body diagram. i put space in between each crate. so concentrating on the 3rd crate, i have horizontal forces of f_k to the left and another leftward force of crate 3 acting on crate 2 (F23). this force of which should be the same as the force of crate 2 acting on crate 3 (F32). so i got this equation of:

    -F23 - f_k = m3*a_x therefore: F32 which equals F23 would then be:
    F32 = -m3*a_x - f_k

    BUT, i totally see what you are saying; and now i am confused as to how my equations don't work (giving me the wrong sign). 'cause.....are you saying there is no leftward force from crate 3 acting on crate 2. and, if there is, is it not equal to the force i am asked for which is the force on crate 3 from crate 2? oh boy, i hope my rambling isn't giving you a headache. i guess the point is, the answer is 147 N. i guess it's just a matter of me getting my free body diagrams right and getting my signs right.

    nonetheless: thank you so much, doc al!
     
  11. Mar 4, 2007 #10

    Doc Al

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    Staff: Mentor

    Why put space between them? They are touching!
    When analyzing the force on the 3rd crate, all you care about are forces on the 3rd crate. You don't care about the leftward force that the 3rd crate exerts on the 2nd crate, but you do care about the rightward force that the 2nd crate exerts on the 3rd.
    Same magnitude, but different sign. F23 = -F32.
    Oddly enough, your first equation is correct: -F23 - f_k = m3*a
    But you really should have written it as: F32 - f_k = m3*a
     
  12. Mar 4, 2007 #11
    :approve: okay

    i drew it that way 'cause that's what my prof does. hmmm, bad habit i guess. tsk tsk.

    so since F32 and F23 have different signs, that completely makes sense then! thank you really!!!
     
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