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Friction and Forces

  1. Nov 26, 2007 #1
    1. The problem statement, all variables and given/known data

    A 200-kg crate is pushes horizontally with a force of 700 N. If the coefficient of friction is 0.20, calculate the acceleration of the crate.

    2. Relevant equations

    Well, at the moment I'm just wondering if I used the right equations, and if not then a heads up on where I went wrong :eek:

    Would I start with Ffr = u x Fn

    and then proceed to a = Fnet / m?

    3. The attempt at a solution

    Fn = 200kg x 9.8 = 1960

    Ffr = u x Fn, so 1960 x .20 = 392.

    Ffr = 392.

    a = Fnet / m,

    Fnet = Fapplied + Ffr = 700N + 392 = 1092
    [Removed Fg and Fn since they cross out as the block is not falling or rising]

    Fnet / m = 1092 / 200 = 5.46 m/s


    = = = =

    I have another question that I'm currently stuck on but have yet to totally give it a try, so I will post the question and update it accordingly.

    1. The problem statement, all variables and given/known data

    Safety engineers estimate that an elevator can hold 20 persons of 75-kg average mass. The eleveator itself has a mass of 500 kg. Tensile strength tests show that the cable supporting the elevator can tolerate a maximum force of 2.96 x 10^4 N. What is the greatest acceleration the elevator's motor can produce without breaking the cable.

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 26, 2007 #2

    rl.bhat

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    Homework Helper

    Fnet = Fapplied + Ffr = 700N + 392 = 1092
    Fapplied and Ffr act in the opposite direction. Therefore Fnet = Fapplied - Ffr
     
  4. Nov 26, 2007 #3
    Ahh, I always throught it was Fapplied = Fnet - Ffr with Fnet = Fapplied + Ffr.. So it would be 700 - 392 = 308, 308 / 200 = 1.54 m/s. Alright, thanks for the help :) I'll probably come back tomorrow for the next question.
     
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