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Homework Help: Physics Dynamic problem

  1. May 28, 2007 #1
    1. The problem statement, all variables and given/known data
    we want to move an object with mass m=100kg which is still on a rough surface with a force F=800N. between the surface and the object there's a [tex]\mu_s=1[/tex]. Prove that the object can be moved if the force is applied with a 30° angle with the surface.

    2. Relevant equations

    F=m*a (N's 2nd law)

    3. The attempt at a solution
    friction: A=-mu *N=-100*10=-1000N
    F_rx=m*a_x ==> a_x=(-mu*N + F*cos30°)/m=-3.07 m/(s*s)

    F_ry=m*a_y ==> a_y=(F*sin30)/m=4

    F_r=sqrt( F_rx^2 +F_ry^2)

    But I get that A>F_r (result of all forces) so the body doesn't move.

    Where am I wrong?

    Attached Files:

  2. jcsd
  3. May 28, 2007 #2
    someone please help me!!
  4. May 28, 2007 #3
    Hello.I cant follow your answer what is N? But friction is Mu times the reaction which is mg - sin 30. F cos 30 is in fact greater than mu times mg- sin 30.Therefore object moves.Hope it is of help.
  5. May 28, 2007 #4
    N is the normal force to the surface, what you called reaction. sorry didn't get your suggestion: why is Normal force (N)

    [tex]N= mg - \sin 30[/tex] ?

    I thought N was cancelled by [tex]\vec P= - m \cdot \vec g[/tex]

    I know that F has two components, x and y

    [tex]F_x= |\vec F| \cdot \cos 30 [/tex]

    [tex]F_y= |\vec F| \cdot \sin 30 [/tex]

    but N has only one component: along the positive y
    Last edited: May 28, 2007
  6. May 28, 2007 #5
    This is what I figure out of the forces applied to the object. (see the attachment)

    Attached Files:

    Last edited: May 28, 2007
  7. May 28, 2007 #6

    Doc Al

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    Staff: Mentor

    You must solve for the normal force by considering the vertical components of all forces acting on the object: They must sum to zero. In your diagram it looks as if you just assume the normal force equals mg, which is not true in this case, since other forces act vertically.
  8. May 29, 2007 #7
    I spoke to my teacher and I understood my mistake.
    I was thinking that the reaction force to the surface was ALWAYS only "m∙g". Actually the normal force varies, it's not always "m∙g" or, better, it's not just that.
    The gravity force balances the normal force AND the F_y=F∙sin30°

    mg= N+ F∙sin30° → N= mg - F∙sin30°.

    Once found N, I can calculate the friction force A=µ∙N.
    I must then check whether it is A<F∙cos(30°). If so the body moves :)
    Last edited: May 29, 2007
  9. May 29, 2007 #8
    The normal reaction force of the surfce on the object is equal and opposite to the force exerted by the object on the surface which in this case is not mg but mg - F sin 30.
  10. May 29, 2007 #9
    so you are saying N=mg - F∙sin(30°) is not right?
  11. May 29, 2007 #10
    No. im saying its right .... but u said that the normal force varies i wanted to add that it is equal n opposite to the force that the object exerts on the surface. So u knew that i think?
  12. May 30, 2007 #11
    ok, thank you very much!
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