- #1
phys101
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- 0
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 = VoT + 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)
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 = VoT + 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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