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

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AI Thread Summary
The discussion focuses on calculating the reaction force on a box being pulled up a ramp at a 27-degree angle, with a mass of 50 kg and a coefficient of kinetic friction of 0.3. Participants explore the necessary equations and components of forces acting on the box, including gravitational force, frictional force, and the applied horizontal force. There is confusion regarding the correct angle to use for resolving forces, particularly how to incorporate the ramp's angle with the horizontal force applied. The importance of determining whether the box is moving at constant velocity is also highlighted, as it affects the net force calculations. Understanding these forces and their components is crucial for accurately solving the problem.
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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
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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 = VoT + 1/2AT ** Vf = Vo + AT ** V²F = V²o + 2A ∆X ** ∆X = [(Vo + Vf)/2]T


3. The Attempt at a Solution

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problem diagram work.png
problem diagram working part 2.png or
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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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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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