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Normal and Friction Forces

  1. Jun 15, 2008 #1
    A 16.2 kg object is pulled along a frictionless horizontal surface by a horizontal net force of 10.2N. What is the normal force acting on the object?

    Fn= 16.2*9.8

    I am sure do i have to ignore the 10.2N force....
    Thank you for helping
  2. jcsd
  3. Jun 15, 2008 #2
    Sorry I'm not a physics major rather a struggling student myself, but this was last section for me, so I can try to help perhaps.. I think if you can say that Voy = 0 and Vfy=0 since there is no vertical displacement the normal force must be equal to the weight of the object, therefor: MG = Fn = (16.2)(9.8), unless there is a slope that the object is on affecting the acceleration you should keep the vertical / horizontal components separate.
  4. Jun 15, 2008 #3
    Yes, that's correct.

    Since, [tex]F_{n,y}+F_{w,y}+F_{y}[/tex](where F_n is the normal force and F_w is the weight; *y means in the y-component form) and we know that [tex]F_{y}=0[/tex](the y-component of the horizontal force) since the angle [tex]\theta=0[/tex] and [tex]F_{y}=Fsin(\theta)=0 [/tex], then the normal force would just be equal to the weight.
    Last edited: Jun 15, 2008
  5. Jun 16, 2008 #4
    Well done.
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