Normal and Friction Forces

  • Thread starter blue__boy
  • Start date
  • #1
8
0
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?

Answer:
Fn= 16.2*9.8
=158.8

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

Answers and Replies

  • #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.
 
  • #3
238
0
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:
  • #4
335
0
Well done.
 

Related Threads on Normal and Friction Forces

  • Last Post
Replies
7
Views
725
  • Last Post
Replies
5
Views
5K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
5
Views
2K
Replies
3
Views
2K
  • Last Post
Replies
2
Views
3K
Replies
1
Views
554
Replies
1
Views
12K
  • Last Post
Replies
7
Views
2K
Replies
2
Views
2K
Top