1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.


    Thanks.
     
  2. jcsd
  3. Mar 26, 2012 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    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).

    ehild
     
  4. Mar 26, 2012 #3

    NascentOxygen

    User Avatar

    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 :).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook