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Determining max horizontal force using friction coefficient

  1. Nov 5, 2012 #1
    1. The problem statement, all variables and given/known data
    A 0.5 kg wooden block is placed on top of a 1.0kg block. The coefficient of static friction between the two blocks is 0.35. The coefficient of kinetic friction between the lower block and the level table is 0.2. What is the maximum horizontal force that can be applied to the lower block without the upper block slipping? The correct answer on the sheet is 8.1N


    2. Relevant equations
    aΔt=v2-v1
    d= 1/2(v1 + v2)∆t
    d=v1Δt + 1/2aΔt^2
    v2^2=v1^1 + 2ad
    F=ma
    μ=Ff/Fn


    3. The attempt at a solution

    (9.8)(0.2)=Ff of bottom 1kg box
    1.96N=Ff

    (0.5)(0.35)=Ff for top 0.5 kg box
    0.175N=Ff

    Unsure of the next step. I assumed the force would be equal to the frictional force of the bottom box but it is not.
     
  2. jcsd
  3. Nov 5, 2012 #2

    Nugatory

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    Staff: Mentor

    There is only one force acting on the upper block. You could try considering what the maximum value of that force could be, and hence what the maximum possible acceleration of the upper block is going to be.
     
  4. Nov 5, 2012 #3
    The maxiumum value of force is the same as the Ff (0.175 as shown above) If I calculate accelertion 0.175N/0.5kg=a i get 0.35m/s^2

    If I put that acceleration to calculate the max force of the other block (1kg)(0.35m/s2)=F i get 0.35N again....which is wrong
     
  5. Nov 5, 2012 #4

    PhanthomJay

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    Science Advisor
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    Gold Member

    You first should correct your value for the friction force on the top block.....you forgot to multiply by g.
    Then for the friction force on the bottom block from the table, you must use the normal force, not the weight. And when applying newton 2 to the system, you must use the net external force.
     
  6. Nov 5, 2012 #5

    Nugatory

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    Staff: Mentor

    Have you included all the forces acting on the lower block to get it to accelerate at particular rate? There's the frictional force from the table to consider... And if friction is accelerating the upper block as well, there's an equal and opposite force to consider (or you could just consider that as long as the upper block isn't sliding, the force applied to teh lower block is accelerating both blocks).
     
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