Find the acceleration of the box.

  • Thread starter rickylin89
  • Start date
  • #1
16
0
I really need some help with this problem. PLEASE HELP!!

Here it is:

You are pushing a box uphill the hill has a pitch of 15 degrees above the horizontal, and the static and kinetic friction coefficients between the box and the hill are 0.45 and 0.35. The box has a mass of 40 kg. You push horizontally with a push of 180 N, and the box is initially moving uphill. Find the acceleration of the box.


ANY HELP WOULD BE GREAT!!!
 

Answers and Replies

  • #2
16
0
I really need some help with this problem. PLEASE HELP!!

Here it is:

You are pushing a box uphill the hill has a pitch of 15 degrees above the horizontal, and the static and kinetic friction coefficients between the box and the hill are 0.45 and 0.35. The box has a mass of 40 kg. You push horizontally with a push of 180 N, and the box is initially moving uphill. Find the acceleration of the box.


ANY HELP WOULD BE GREAT!!!
 
  • #3
tiny-tim
Science Advisor
Homework Helper
25,836
251
Welcome to PF!

Hi rickylin89! Welcome to PF! :wink:

Show us what you've tried, and where you're stuck, and then we'll know how to help. :smile:

(but please don't double-post :frown:)
 
  • #4
16
0
Honestly, I don't even know how to start the problem. I really don't know what to do lol.
 
  • #5
2,358
683
Perhaps you could begin by drawing a free body diagram and summing forces parallel and perpendicular to the inclined surface.
 
  • #6
16
0
I'm not really sure what the free body diagram would look like.
 
  • #7
2,358
683
The free body diagram will show the box and all of the forces that act on the box. These include the force of gravity (the weight), the normal reaction force from the surface, the friction force parallel to the surface, and the horizontal applied force of 180 N. Once you have this figure, it will help you see what forces act in each direction so that you can properly write the force sums for Newton's second law.
 
  • #8
16
0
Alright I drew the free body diagram. I believe that the net force is the up force, the weight is the down force, the force of friction is to the leff of the box, and the force of the push is to the right of the box. The force of friction and force of the push are both at an angle though. Does this seem right?
 
  • #9
2,358
683
Who (what?) is this fellow "net" force? Where does he (it) come from? (Do you know what the term "net force" means? It really does not sound like you do, and that is something you need to understand. Please look up the definition.)

The problem statement said, "You push horizontally," but now you tell me, "The ...force of the push are both at an angle..." Which way is it?

You are correct in saying that the weight acts down. I think you still have some thinking to do about the direction of the other forces, and just what they come from.

A term that you ought to know is "normal reaction." It describes the reaction force normal (perpendicular) to a surface. It is a useful term.
 
  • #10
16
0
I know that the force of the push is equal to 180sin(15) but I don't understand how to incorporate the force of friction.
 
  • #11
2,358
683
I know that the force of the push is equal to 180sin(15) but I don't understand how to incorporate the force of friction.

Unless you have changed the problem statement, you don't know this. The original problem statement said that the push was 180 N.

You still need to sort out your FBD, and until you do that, you are not going to get very far. Please don't try to cut corners. It is costing you time.
 
  • #12
16
0
I don't understand what you are asking me to do.
 
  • #13
16
0
What I meant to say was.... I know that the NORMAL force is mgcos(15). I believe that is right because I drew out the components. I am just sure if it is cos(15) or sin(15)?
 
  • #14
tiny-tim
Science Advisor
Homework Helper
25,836
251
What I meant to say was.... I know that the NORMAL force is mgcos(15). I believe that is right because I drew out the components. I am just sure if it is cos(15) or sin(15)?

Hi rickylin89! :smile:

:rolleyes: hmm … there are two ways to work this out …

i] suppose the angle was 0º instead of 15º … would the normal force be mgcos(0) or mgsin(0)? :wink:

ii] components are always cos of the angle between the force and the direction … sin only comes in when for example you've only defined one angle, θ say, and you need to use it for two perpendicular directions … then it'll be cos(θ) and cos(90º - θ), and we write the latter as sin(θ)! :smile:
 
  • #15
16
0
So then it's mgcos(15)?
 

Related Threads on Find the acceleration of the box.

  • Last Post
Replies
23
Views
3K
  • Last Post
Replies
2
Views
820
  • Last Post
Replies
5
Views
2K
Replies
2
Views
745
  • Last Post
Replies
15
Views
2K
Replies
3
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
13
Views
1K
Replies
3
Views
1K
Top