Homework Help: Box being pulled up a ramp by a horizontal force

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

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• problem diagram working part 2.png
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Last edited by a moderator: May 3, 2017
2. Nov 26, 2007

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)

3. Nov 26, 2007

phys101

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

gonna try again

4. Nov 26, 2007

phys101

changed the jpegs into pngs

5. Nov 26, 2007

phys101

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

6. Nov 26, 2007

rl.bhat

It shows Attachments Pending Approval

7. Nov 26, 2007

phys101

lol, they dont have viruses. And both the jpegs and the pngs worked/working for me. But that might just be because I'm the poster?

Last edited: Nov 26, 2007
8. Nov 26, 2007

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 [Broken]

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

phys101

10. Nov 26, 2007

phys101

11. Nov 26, 2007

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.

12. Nov 26, 2007

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.