Pulling a box up a ramp - Newton's second law

In summary, when finding the sum of forces in the x and y directions for a box being pulled up a ramp with a force F, you must take into account the force of gravity projected onto each direction. The force of gravity in the x direction will be mg sin theta and in the y direction it will be mg cos theta. Additionally, you must use the correct sign for the acceleration due to gravity, which depends on how you define your vectors. It is important to subtract forces that point in opposite directions to accurately calculate the net force.
  • #1
mybrohshi5
365
0

Homework Statement



A box is being pulled up a ramp with a force F. The ramp has a 30 degree incline with the horizontal. write the sum of the forces in the x direction and y direction. Make the x and y coordinate system be parallel and perpendicular with the ramp.

The Attempt at a Solution



Ok so my sum of forces in the x direction (parallel to the ramp) is

F_pull - F_gx = ma

Force of the pull minus force of gravity on the box in the x direction equals mass times accel.


My sum of forces in the y direction (perp. to the ramp) is

n = mg

Normal force equals mass times accel due to gravity


I am having trouble grasping this. Please help me understand :)

Thank you
 
Physics news on Phys.org
  • #2
Not all of the force of gravity is acting in the x direction. Nor all in the y direction. You have to project the force of gravity onto the x and y directions to find how much of it is in each direction. You should get one of them to be mgsin(theta) and the other to be mgcos(theta). Figure out which is which. =)
 
  • #3
Is there friction? You need to draw a free body diagram of the box, set your coordinates so that x is parallel to the ramp and y is perpendicular to the box. The draw the forces on the box. What is the force of gravity in the x direction? What about the y direction? Then just sum them up in each direction using Newton's 2nd Law.
 
  • #4
I understand that the x force of gravity will be mg sin theta and the y will be mg cos theta

I did draw a FBD.

Mainly what i am confused about is adding or subtracting.

So for the sum in the x is it

F_pull + mgsin(theta) = ma

or

F_pull - mgsin(theta) = maand there is no friction :)
 
  • #5
then the forces in the y direction just equal zero right?

cause

N - Fg_y = 0

mgcos(theta) - mgcos(theta) = 0
 
  • #6
This depends on how you define your vectors. For example if you use g=-9.8m/s^2 or g=+9.8m/s^2.

Make sure that if the vectors point opposite, then in the end you are subtracting! It would make no sense if vectors point opposite but somehow the force gets stronger!
 
  • #7
so if on my fbd i have

Fg_x going down the ramp and F_pull going up the ramp i should have

F_pull - Fg_x = ma

then use 9.8m/s^2 for g?
 

1. How does Newton's second law relate to pulling a box up a ramp?

Newton's second law, also known as the law of acceleration, states that the force applied to an object is directly proportional to its mass and acceleration. In the case of pulling a box up a ramp, the force of gravity pulling the box downwards is balanced by the force exerted by the person pulling the box upwards. This force must be greater than the force of gravity in order to overcome the weight of the box and accelerate it up the ramp.

2. What factors influence the force required to pull a box up a ramp?

The force required to pull a box up a ramp is influenced by the mass of the box, the angle of the ramp, and the force of gravity. The greater the mass of the box, the greater the force needed to overcome its weight. As the angle of the ramp increases, the force required to pull the box up also increases. Finally, the force of gravity will always pull the box downwards, requiring a greater force to pull it up the ramp.

3. Does the length of the ramp affect the force needed to pull a box up?

Yes, the length of the ramp can affect the force needed to pull a box up. A longer ramp will require a smaller force to pull the box up, as the incline is less steep. This is due to the fact that the longer ramp allows the person pulling the box to exert the force over a longer distance, reducing the amount of force needed to overcome the weight of the box.

4. How does friction play a role in pulling a box up a ramp?

Friction, the force that resists motion between two surfaces in contact, can make it more difficult to pull a box up a ramp. Friction can be reduced by using a smoother surface for the ramp or by using lubricants. However, some amount of friction is necessary to prevent the box from sliding down the ramp too quickly. The force of friction must be taken into account when calculating the force needed to pull the box up the ramp.

5. Is there a limit to the force that can be exerted when pulling a box up a ramp?

Yes, there is a limit to the force that can be exerted when pulling a box up a ramp. This limit is determined by the maximum force of friction between the box and the ramp, as well as the strength and ability of the person pulling the box. If the force exerted exceeds this limit, the box may start to slide down the ramp or the person may become fatigued and unable to continue pulling with the same force.

Similar threads

  • Introductory Physics Homework Help
Replies
11
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
716
  • Introductory Physics Homework Help
Replies
3
Views
959
  • Introductory Physics Homework Help
2
Replies
56
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
916
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
13
Views
963
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
2K
Replies
44
Views
1K
Back
Top