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Using Newtons Law: Friction

  1. Jun 10, 2012 #1
    1. The problem statement, all variables and given/known data

    The static friction between the two blocks is .47 and kinetic friction between the block with mass 9.4 kg and the horizontal surface is 0.21.The acceleration of gravity is 9.8 m/s^2 What is the minimum force F which must be exerted on the 9.4 kg block in order that the 0.8 kg block does not fall? Answer in units of N

    diagram: attached

    2. Relevant equations

    ƩF=m*a
    Fμ = μ*Fn

    3. The attempt at a solution

    Free Body Diagram: attached

    F(net on big block) = F - (.21*m*g)

    F(net on small block) = F - (.47*m*a)

    Now I made an assumption that F(net on big block) = F(net on small block)

    F - (.47*m*a) = F - (.21*m*g)
    0 = - (.21*9.4*9.8) + .47*(.8)*a
    19.3452 = .376a .... a = 51.45 m/s^2 ?

    F(net on small block) = m*a - (.47*m*a) = (.8*51.45) - (.47*.8*51.45) = 21.815 N

    I went through this with a TA who told me we could make that assumption, but seeing I got the answer wrong I am quite confused at this point :( Is my approach somewhat correct? I do believe this is the right section to post this if it isn't I am sorry! this problem has been giving me an headache for quite sometime some clarification would be welcomed :)
     

    Attached Files:

  2. jcsd
  3. Jun 10, 2012 #2

    HallsofIvy

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    Since the static friction between the two blocks is .47, the vertical force on the smaller block, holding it up, is .47f where f is the force the larger block is applying to it. That f is, in turn, .8a where a is the acceleration of the larger block. If the force applied to the larger block is F, then a= F/9.4. So the vertical force on the smaller block is .47(.8(F/9.4). In order that the small block not fall, that must be larger than the gravitational force on the smaller block, 0.8g.
     
  4. Jun 10, 2012 #3
    First you have to find the force needed to hold the 0.8kg block from falling down.
    When this force is exerted on the small block, what is the acceleration cause by the force.
    This acceleratioon must be the same for all the bodies, they are moving as one piece.

    Edit: acceleration is not needed in this problem
     
    Last edited: Jun 10, 2012
  5. Jun 10, 2012 #4
    @HallsofIvy .8(9.8) > .47(.8(F/9.4)) therefore (if I did the calculations correctly) 196 > F How does this help me solve for F ? I am probably misunderstanding your point...

    @azizlwl Wouldn't the force needed to hold up the block be F(net on small block) = F - (.47*m2*a) and F(net) = 0 ? seeing we don't want it to fall ? Then I suppose I could take the information provided by HallsofIvy that a = F/9.4 and plug that into the above equation to get F = (.47*.8*(F/9.4)) ? but I don't see how that would help either...

    sorry if I am not understanding your responses correctly :S
     
  6. Jun 10, 2012 #5
    to balance the small block's weight: f=mg
    f=μN

    μN=mg
    N=mg/μ
    ma=mg/μ
    a=g/u
    Since it is a static friction, the friction value from zero to maximum μ. Since μ is at maximum, then a is at minimum.

    this is the acceleration of the combine small and big block.
    Now find the force needed to push a block(small + big) against friction and result in minimum acceleration as above.
    Thus minimum force.
     
    Last edited: Jun 10, 2012
  7. Jun 10, 2012 #6
    From my understanding the following should be correct?

    F(normal) = m1*g therefore F(friction) = μ(k) * m1 * g

    ƩF = F(applied) - F(friction) = (m1 + m2)*a
    F = (m1+m2)*(g/μ(s)) + μ(k)*m1*g

    Although this gives me a very large applied force...
    F(friction) isn't μ(k) * (m1 + m2) * g is it?
     
  8. Jun 10, 2012 #7
    We also need to know the concept behind the question on minimum force. Any value of force we get should also verified it is a mininmum value.

    F=(m1+m2)μkg +Nm2

    Edit: Correction, acceleration value is not needed in this problem. 3rd. Newton's law for the interaction between small and big block.
     
    Last edited: Jun 10, 2012
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