# Homework Help: Box on a ramp (resubmit)

1. Nov 26, 2007

### phys101

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

A box (mass = 50.0 kg) is being pulled up a ramp at an angle of (27 ⁰) by a horizontal force, (F→).
Given that the coefficient of kinetic friction for the ramp is 0.3, find the force, (F→) on the box.

What is the reaction force on the box due to the tramp? (Identify type of force, what causes the force, what the force is acting on, and the direction that force is acting).

(DIAGRAM: https://www.physicsforums.com/attachment_browser.php [Broken]
problem diagram.png or
http://img517.imageshack.us/my.php?i...problemys8.jpg)

2. Relevant equations

Key: ∆X = distance X direction
Vo = Initial Velocity
Vf = Final Velocity
A = Acceleration
T = Time
*Possible equations given to solve the Problem*
∆X = VoT + 1/2AT ** Vf = Vo + AT ** V²F = V²o + 2A ∆X ** ∆X = [(Vo + Vf)/2]T

3. The attempt at a solution

(DIAGRAM: see
https://www.physicsforums.com/attachment_browser.php [Broken]
problem diagram work.png
problem diagram working part 2.png or
http://img517.imageshack.us/my.php?i...problemys8.jpg
http://img517.imageshack.us/my.php?i...icsworkri9.jpg
http://img147.imageshack.us/my.php?i...rkingpaos9.png)

The X and Y axis have been rotated to accommodate the new angle 27 ⁰

μ (coefficient of kinetic friction)= .3
Ff = (μ)(Fn)
Ff = (.3)(50.0kg)(9.81 m/s^2)
Ff = 147.15 N (Newtons)
Fn = ma (mass)(acceleration)
= (50.0kg)(-(-9.81m/s^2)
= 490.5 N
Fgy = (50.0kg)(-9.81m/s^2)
= -490.5 N
What I think I need…
Fgx = (50.0kg)(9.81m/s^2)(sin ? ⁰)
My difficulty is (sin ? ⁰)…

Because (F →) is horizontal instead
of being parallel with the X axis, how do I find the proper sin angle?
63 ⁰? Or 63 ⁰ + 27 ⁰? Or 63 ⁰ - 27 ⁰?...

If I can find the proper sin angle would I then equate (F →) as the Fnet (net force) by adding
Fgy + Fgx + Ff?

How are the given kinematic equations involved in finding the solution? (if at all)

Last edited by a moderator: May 3, 2017
2. Nov 26, 2007

### phys101

imageshack seems to have crashed and im giving up

3. Nov 26, 2007

### Astronuc

Staff Emeritus
See this example, which uses a pushing as opposed to pulling force.
http://hyperphysics.phy-astr.gsu.edu/hbase/faia.html#c2

The angle for the ramp is conventionally with respect to horizontal. Resolve all the forces into components parallel and normal to the ramp.

Also, is the box moving at constant velocity? Otherwise the solution could be any force greater to or equal to the forces (friction and weight component) parallel with the ramp surface.