# Normal and Friction Forces

1. Jun 15, 2008

### blue__boy

Question:
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
=158.8

I am sure do i have to ignore the 10.2N force....
Thank you for helping

2. Jun 15, 2008

### zooboodoo11

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.

3. Jun 15, 2008

### konthelion

Yes, that's correct.

Since, $$F_{n,y}+F_{w,y}+F_{y}$$(where F_n is the normal force and F_w is the weight; *y means in the y-component form) and we know that $$F_{y}=0$$(the y-component of the horizontal force) since the angle $$\theta=0$$ and $$F_{y}=Fsin(\theta)=0$$, then the normal force would just be equal to the weight.

Last edited: Jun 15, 2008
4. Jun 16, 2008

Well done.