# Normal and Friction Forces

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

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.