Box on a ramp: What is the reaction force on the box due to the ramp?

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The discussion focuses on calculating the reaction force on a box being pulled up a ramp at a 27° angle, with a mass of 50.0 kg and a coefficient of kinetic friction of 0.3. The frictional force (Ff) is calculated as 147.15 N, while the normal force (Fn) is determined to be 490.5 N. The participant seeks clarification on how to resolve the forces, particularly the gravitational force components and the correct angle to use for the sine function in their calculations. The discussion emphasizes the importance of resolving forces into components parallel and normal to the ramp.

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1. Homework Statement

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
problem diagram.png or
http://img517.imageshack.us/my.php?i...problemys8.jpg)

2. Homework Equations

Key: ∆X = distance X direction
Vo = Initial Velocity
Vf = Final Velocity
A = Acceleration
T = Time
*Possible equations given to solve the Problem*
∆X = volt + 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
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)
 
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imageshack seems to have crashed and I am giving up
 
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.
 

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