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Homework Help: Force - A cart pushing a block

  1. Sep 21, 2010 #1
    1. The problem statement, all variables and given/known data

    A cart of mass mA = 5.7 kg is pushed forward by a horizontal force F. A block of mass mB = 0.76 kg is in turn pushed forward by the cart as show on the picture:

    (The picture shows 9.8m/s^2 pointing down, Force coming from left to right point at the cart which weighs 5.7 kg which then has a block (not of the same size, smaller) weighing 0.76 kg positioned in the middle of its side. They have a friction constant of 0.63)

    If the cart and the block accelerate forward fast enough, the friction force between the block and the cart would keep the block suspended above the floor without falling down. Given g = 9.8 m/s2 and the static friction coefficient μs = 0.63 between the block and the cart; the floor is horizontal and there is no friction between the cart and the floor. Calculate the minimal force F on the cart that would keep the block from falling down. Answer in units of N.

    2. Relevant equations
    Netforce = ma

    3. The attempt at a solution
    I have two free body diagrams but I think my normal force is incorrect. On the 5.7 cart, I have Force point from left to right, P21 point from right to left (the force from the .76 block on the 5.7 block), friction point up and w pointing down. I have normal force point up but I feel that it might be incorrect because friction is also present in that direction.
    For my second, I have P12 from left to right (force from 5.7 block to .76 block), friction pointing up and w pointing down. With this normal force also pointing up but I have the same issue.

    So for the 5.7 block, I have xnetforce = F - P21 = 5.7ax. For the ynetforce = n1+f1-w1=5.7ay.
    For .76, xnetforce = P12=.76ax. ynetforce = n2+f2-w2=.76ay.

    Is my normal force wrong or am I totally missing how I can solve this?
  2. jcsd
  3. Sep 21, 2010 #2


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    The main thing is that the friction force is just enough to prevent gravity from pulling block B down, so start with Ff = Fg.
    The normal force is the force pressing A against B; the same force that causes B to accelerate. If you put in the details for Ff and Fg, you should be able to find the acceleration.

    After that, you still have to find the force necessary to accelerate the cart plus block.

    Good luck!
  4. Sep 22, 2010 #3
    I don't understand how I have enough information to get what acceleration is.

    For my formulas I have


    If friction is equal to normal force times the coefficient (0.63 in this case) but you're saying friction equals the force of gravity (which I assume to mean friction equals m*9.8), then my normal forces would be n1 = 88.66666 and n2=11.82222.
    Even with that information, if I add my two Fnetx together, I get:
    (Change the second sentence to read P21+n2=-m2ax, because P12=-P21, then to make the P21 cancel out, you times that sentence by -1)


    Putting in the numbers I used, I'd still have two variables. F - 76.84444377 = .494ax.

    You said that normal force equaled the force pressing A against B, so would that mean P12 equals n1 or n2?
  5. Sep 22, 2010 #4


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    I can't seem to follow
    What is P21? What is n1? Units of force on the right, but units of force times mass on the left?

    I would start with Ff = Fg
    Then fill in some details: μ*Fn = mg
    There is only one acceleration; the pair of blocks is staying together as they accelerate. Fn is the push of A on B, making B accelerate at "a", so Fn = ma where m is the mass of B. Looks like the mass cancels out, reflecting the fact that a bigger mass (say two blocks one above the other) would behave the same way as one.
  6. Sep 22, 2010 #5
    Okay, using what you said I got my acceleration is 116.6666579. Which means my force would be F=ma, where m = 5.7 and acceleration equals 116.6666579, correct? So I got F = 664.99995 N.

    I just want to check this before I put it in my online homework.
  7. Sep 22, 2010 #6


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    I got about 7 times less for the acceleration. If you show your work, we should be able to sort out our differences.
  8. Sep 22, 2010 #7
    So Fn=ma.

    For mass do we use the two blocks combined, meaning 6.46. If so, then acceleration=n/m, where n = 88.6666 and m = 6.46. So a = 13.72547988.

    Is this the acceleration you got? If so then F = ma = 78.23523532 N
  9. Sep 22, 2010 #8


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    Sorry, I have such difficulty following your work. I have no idea what "n" is and 88.6666 doesn't ring a bell either. We agree on
    Ff = Fg
    μ*Fn = mg
    μ*ma = mg
    where the m in ma is the mass of block B as is the m in mg.
    Looks like your next step is to divide both sides by m and solve for a. Your F = ma for the last step looks good (with the new value for a).
  10. Sep 22, 2010 #9
    Okay, so..

    μ*Fn = mg

    μ*ma = mg

    f = 6.46 (15.55555)
    F = 100.48887 N

  11. Sep 22, 2010 #10


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    You are going to numbers way too soon! Your answer is correct, though.
    Ff = Fg
    μ*Fn = mg
    μ*ma = mg
    μ*a = g
    a = g/μ = 9.8/.63 = 15.6
    The mass does not matter! Number not needed. You can actually see this without formulas if you imagine mass B as being made up of many tiny particles. Each particle experiences the same tendency to fall and to be held by friction. So it doesn't matter how many particles there are; the whole thing will behave the same way. If that doesn't make sense, look up Galileo's explanation of why a big mass falls with the same acceleration as a small mass. It was in a play called "Two New Sciences".

    Sorry to preach, but the idea that the mass doesn't affect the result can be far more important than the numerical answer.
  12. Sep 23, 2010 #11
    Don't apologize for preaching! Haha. Physics isn't my thing but I'm trying to learn the ideas behind it. So, yes, thank you.
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