# Need help with newtons laws. finding normal force.

1. Oct 5, 2011

### ThatStormy

1. The problem statement, all variables and given/known data
It includes a diagram so i just took a screenshot of the question.

I don't undertand how to do this problem
Thanks

Last edited: Oct 5, 2011
2. Oct 5, 2011

### iRaid

What does the normal force equal?

EDIT: you should probably draw a free body diagram to start also.

3. Oct 5, 2011

### Rayquesto

you can compute the normal force knowing two things: trigonometry and the laws of natural states of motion.

4. Oct 5, 2011

### ThatStormy

Yes i know but how?

5. Oct 5, 2011

### iRaid

$F_{N}=mgcos(\theta)$ is what you would use. But really, just draw a FBD and figure out this yourself with the trig.

6. Oct 5, 2011

### ThatStormy

Yeah i got 28.411 but it is wrong...

7. Oct 5, 2011

28.412?

8. Oct 5, 2011

### Rayquesto

It's hard to see how you can get the normal force vector, but you can see it if you draw vectors large enough and spend enough time until you finally see the solution. The natural states of motion are rest and uniform speed which makes the net force 0.

9. Oct 5, 2011

### ThatStormy

yeah its wrong.

10. Oct 5, 2011

### iRaid

28N is probably the answer, if not then I don't know why it's marking it wrong and for the other 1 you just do $F_{g}sin\theta$

11. Oct 5, 2011

### Rayquesto

the other one is the sintheta version because in static equalibrium the action is going to be equal to the reaction and in this case its the gravity in the x direction of the block. so, 0=Fpush-mgsin45.

12. Oct 5, 2011

### ThatStormy

56.824 N (I multiplyed 2 x 28.412 in a desperate attempt to guess)
40.180 N (4.1 x 9.8 and also 28.412^2 + 28.412^2 = 40.180^2)

I have no idea why or how these worked lmao

13. Oct 14, 2011

### ErwinMoses

N=mg(cosθ) +Fcosθ.......(1)

Fsinθ=mgsinθ.......(2)
i.e,F=mg=4.1 x 9.8=40.18
so,F=40.18

from (1),N=mgcosθ +mgcosθ [F=mg]
i.e,N=2mgcosθ=2 x 4.1 x 9.8 x cos 45°=56.83
so, N=56.83