1. The problem statement, all variables and given/known data(adsbygoogle = window.adsbygoogle || []).push({});

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).

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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

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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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# Homework Help: Box on a ramp (resubmit)

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