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Physics Friction

  1. Nov 10, 2014 #1
    1. The problem statement, all variables and given/known data
    While dragging your 22.3 kg catapult across the floor of your garage, you wonder what coefficient of friction is between the catapult and the concrete. If you accelerate your catapult at 0.76 m/s2 with a force of 79 N at 21°, find:

    1. Find the components.
    Fx=
    Fy=
    2. Force Normal: FN =
    3. What is the coefficient of friction (mu/μ) between the dry concrete and the wood of your catapult? μ=
    2. Relevant equations
    kinematics2.jpg where Vf= final velocity Vi= initial velocity a= acceleration t= time
    kinematics3.jpg x= distance in the x direction (can also be replaced with y in the y direction)
    kinematics1.jpg
    Fnet= ma (Net force= mass times acceleration)

    Ff= μ FN (Force of friction= mu[coefficient of friction] times Normal Force)
    3. The attempt at a solution

    I have figured out the first three questions correctly;
    Fx= 73.8N using mgcos(theta) (theta is the degree of the incline)
    Fy=28.3N using mgsin(theta) (x and y the postions on a graph)
    Force Normal= 190.2N using Fnormal+Fgravity+Fpull (sin theta)=0

    I can't figure out how to find the mu between the concrete and the wood, I got 4.67 as my answer but the problem is telling me the correct answer is .2968. I know Ffriction is needed but I'm not sure if I'm doing it correctly, since I was not given any equation for it. After finding that I would think just plugging that into the last equation to find mu?
     
    Last edited: Nov 10, 2014
  2. jcsd
  3. Nov 10, 2014 #2

    ehild

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    Homework Helper
    Gold Member

    There is no incline, the floor is horizontal. But you apply a force F which makes the angle of 21°with the horizontal. Fx and Fy are the horizontal and vertical components of your force. Apart of these, the results are correct.

    Show your work in detail. What is the net horizontal force applied on the catapult?
     
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