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Homework Help: Newton problem

  1. Nov 6, 2007 #1
    1. The problem statement, all variables and given/known data

    a person pulls a 40kg car at a 30 degree angle with 230N. If the coefficient of friction is 0.40, with what acceleration does the car move?

    2. Relevant equations

    F =ma
    f = mN

    3. The attempt at a solution

    I draw a FBD

    then I found F sin θ which is 115, and F cos θ which is 199.19. Friction came out to be 392N. I dont know what to do next I need a little bit of help. thanks in advance.
  2. jcsd
  3. Nov 6, 2007 #2
    Well now that you have the applied force and the force of friction, you need to find the parallel force of gravity. Then you use the add up all your forces, equate it to Fnet. Replace Fnet with m*a, divide Fnet with the mass of the car, and you will have the acceleration.
  4. Nov 6, 2007 #3
    I dont think your force of friction is correct.

    392.4N (40kg*9.81) is the weight of the car itself.

    [Fy=0] Fsin30 + FN = 392.4 N

    You can spot this error because your friction force is larger than any other force therefore the object wouldnt even move if that were the case. Unless this was a tricky question.

    Once you have the friction force. Do Fx to find the net force. then use F=ma to find acceleration.
    Last edited: Nov 6, 2007
  5. Nov 6, 2007 #4
    I hope this diagram helps.

    http://img138.imageshack.us/img138/996/vectop0.png [Broken]​
    Last edited by a moderator: May 3, 2017
  6. Nov 6, 2007 #5
    temaire, I dont think the problem states that the car is on an incline but rather being pulled at an angle.
    Last edited: Nov 6, 2007
  7. Nov 6, 2007 #6
    Oh I see, you're right about that. I'am just so used to seeing incline questions worded like that. Sorry for the confustion.
  8. Nov 6, 2007 #7
    thats right, the new value for friction is 156.8, but I still don't understand how to add all the forces. I have friction, the horizontal and vertical components of the force applied, what else do I add?
  9. Nov 6, 2007 #8
    I am getting 110.96 N as the frictional force.

    Well now you have frictional force pushing one way and the force applied Fcos30 pushing the other. subtract them and you will find the net force pushing one way.

    Keep in mind that if the frictional force is more, the object wont accelerate the other way, but rather be still.

    You can then find acceleration by:


    What happened to the vertical force? Well Fy = 0 so the object is in equilibrium vertically. But Fx does not equal 0.
  10. Nov 6, 2007 #9
    thanks, but how did you got 110? I recalclated and had 156.8
  11. Nov 6, 2007 #10
    Fsin30 +FN = mg

    FN= (40)(9.81) -230sin30 = 277.4N

    Ffr = muFN
    Ffr = 0.4(277.4) = 110.96 N


    How are you calculating it?
  12. Nov 6, 2007 #11
    f = m N N=-W
    => 0.40(392N) = 156.8N
  13. Nov 6, 2007 #12
    This is not correct because the normal force is not 392N because there is a force acting on the object at an angle.

    You have to use [sigmaFy=0] to find FN like I have.
  14. Nov 6, 2007 #13
    ok thanks, so I got 2.2 m/s2 for the acceleration
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