Box being pulled up a ramp by a horizontal force

[phys101]

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: see png attachments or https://www.physicsforums.com/attachment_browser.php or http://img517.imageshack.us/my.php?image=boxrampphysicsproblemys8.jpg)

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

The Attempt at a Solution

(DIAGRAM: see png attachments or https://www.physicsforums.com/attachment_browser.php 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?image=problemdiagramworkingpaos9.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)

 

[rl.bhat]

A box (mass = 50.0 kg) is being pulled up a ramp at an angle of (27 ⁰) by a horizontal force, (F→).
Whether the force is parallel to the base of the inclined plane or parallel to the surface of the inclined plane? ( I can't open the attachment)

 

[phys101]

The jpeg was supposed to give the clearest picture possible to the problem. maybe i loaded it wrong...

gonna try again

 

[phys101]

changed the jpegs into pngs

 

[phys101]

The horizontal force is parallel to the base of the angle (the ground)

 

[rl.bhat]

It shows Attachments Pending Approval

 

[phys101]

lol, they don't have viruses. And both the jpegs and the pngs worked/working for me. But that might just be because I'm the poster?

 

[phys101]

I can also view the pngs from the (files) link at the website main menu (top of webpage)
https://www.physicsforums.com/attachment_browser.php

 

[phys101]

I've even signed up with imageshack for another image viewing possibility

http://img517.imageshack.us/my.php?image=boxrampphysicsproblemys8.jpg
http://img517.imageshack.us/my.php?image=boxrampphysicsworkri9.jpg

 

[phys101]

My attempt to answer the 2nd part of the question.

http://img147.imageshack.us/my.php?image=problemdiagramworkingpaos9.png

 

[rl.bhat]

I got the pictures. Now resolve the force F and Fg into two components, One parellel to the inclined plane and another perpendicular to the inclined plane. They are Fcos(27)-parallel, Fsin(27) Perpendicular. Similarly Fgsin(27)--parallel, Fgcos(27) parpendicular. Add the perpendicular compomnents. It becomes normal reaction, from that find the frictional force,Fr. Fr + Fgsin(27) = total downward force. In equilibrium this must be equal to upward Fcos(27). Now solve for F.

 

[kalok]

Sorry,my english is poor.
Is that box stand at ramp(and it o not muve again) ?
If it is,build a coordinate system .x-axis is parallel to inclined plane.
y-axis is perpendicular to the inclined plane.

Now resolve gravity to two parts
f(gx)=-mgsin27*
f(gy)=-mgcos27*

resolve F to two parts
f(Fx)=Fcos27*
f(Fy)=-Fsin27*

kinetic friction:f(μ)=μ[f(Fy)+f(gy)]

For equilibrium:
f(Fx)+f(μ)+f(gx)=0

f(Fx)+μf(Fy)=-f(gx)-μf(gy)

Fcos27*+0.3(Fsin27*)=0.3(50)(9.8)(cos27*)+50(9.8)(sin27*)

[Because (F →) is horizontal instead
of being parallel with the X axis, how do I find the proper sin angle?]
When I study physics at early.I have same problems as you.But it is not difficult.
To build a coordinate system(Origin is the mass center),one axis parallel to path,another perpendicular.
All the force,must resolve to these tow axises.
I hope it can help you.

PS.Pardon me for my lousy English.