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Homework Help: Forces on block question

  1. Mar 26, 2012 #1
    I have a question on block that I was solving in my book.
    I got the answer right but different procedures for doing it I want to make sure my logic is correct,and I didn't get the answer out of probability.

    Two crates of mass 75kg and 110kg, are in contact and at rest on a horizontal surface. A 620N force is exerted on the 75kg crate. If the coefficient of kinetic friction is 0.15, calculate a)The acceleration of the system.

    Here I solved as follows:-

    Fnet1 = F1 - F12 - Ff1 --> F1 - f12 - Ff1 = (m1)(a);

    Fnet2 = F21 - Ff2 --> F21 - Ff2 = (m2)(a);

    We know since objects are in horizontal plane then normal force is equal to weight of objects.
    Fn1 = (m1)(a); --> Ff1 = (u)(m1)(a);
    Fn2 = (m2)(a); --> Ff2 = (u)(m2)(a);

    Fnet1 -> F1 - F12 - (u)(m1)(a) = (m1)(a);
    Fnet2 -> F21 - (u)(m2)(a) = (m2)(a);

    Adding both equations together we'll get net acceleration on the system.

    F1 - F12 - um1g + F21 - um2g = m1a + m2a;

    F - (u)(m1)(g) - (u)(m2)(g) = (a)(m1 + m2);
    (F - (u)(m1)(g) - (u)(m2)(g)) / (m1 + m2) = a;

    Substituting in for values we'll get the following:-

    a = 1.9 m/s^2 ,which is the same answer as in my book but my book got it as follows:-

    Fx = F1 - Fr = (m1 + m2)(a);

    F1 - Fr = (m1 + m2)(a) -> a = (F1 - Fr) / (m1 + m2);
    Substituting for values he got that answer which is the same as mine.
    So did I do something wrong here or maybe somewhere in my math could be modified to get that same formula.

  2. jcsd
  3. Mar 26, 2012 #2


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    Homework Helper

    Your method is correct. The book might mean the force of friction on Fr and uses the result of a derivation, the same what you applied: The acceleration of the CM of a system of interacting bodies equals the resultant of the external forces divided by the total mass.

    The external forces are the applied force F and the forces of friction (and the normal force and gravity which cancel each other).

  4. Mar 26, 2012 #3


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    Staff: Mentor

    Both methods look correct. Whichever way you view it, the applied force must overcome friction, and then the excess accelerates the combined bodies.
  5. Mar 26, 2012 #4
    We are dealing with center of mass. So total mass is m1+m2.
    The center of mass is accelerating.
  6. Mar 26, 2012 #5
    Thanks guys that makes sense :).
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