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Pulling a box up a ramp  Newton's second law 
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#1
May1010, 07:11 PM

P: 365

1. The problem statement, all variables and given/known data
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. 3. 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 


#2
May1010, 07:28 PM

Sci Advisor
P: 2,823

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
May1010, 07:28 PM

P: 258

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
May1010, 07:58 PM

P: 365

Pulling a box up a ramp  Newton's second law
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) = ma and there is no friction :) 


#5
May1010, 08:04 PM

P: 365

then the forces in the y direction just equal zero right?
cause N  Fg_y = 0 mgcos(theta)  mgcos(theta) = 0 


#6
May1010, 08:04 PM

Sci Advisor
P: 2,823

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
May1010, 08:06 PM

P: 365

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? 


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