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

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In summary, the box is being pulled up a ramp at an angle of 27 ⁰ by a horizontal force, (F→). The coefficient of kinetic friction for the ramp is 0.3, so the force, (F→) on the box is 147.15 Newtons. The reaction force on the box due to the tramp is a normal force of -490.5 Newtons.
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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
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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  • #2
imageshack seems to have crashed and I am giving up
 
  • #3
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.
 

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

1. What is a box on a ramp?

A box on a ramp is a common physics experiment that involves studying the motion of a box on an inclined plane or ramp. The box is placed on the ramp and allowed to slide down due to the force of gravity.

2. What are the factors that affect the motion of a box on a ramp?

The motion of a box on a ramp is affected by various factors such as the angle of the ramp, the mass of the box, the coefficient of friction between the box and the ramp, and the force of gravity.

3. How does the angle of the ramp affect the motion of the box?

The angle of the ramp affects the motion of the box by altering the component of the force of gravity that acts parallel to the ramp. As the angle of the ramp increases, the force of gravity acting down the ramp also increases, causing the box to accelerate faster.

4. What is the significance of the coefficient of friction in a box on a ramp experiment?

The coefficient of friction is the measure of the resistance to motion between two surfaces in contact. In a box on a ramp experiment, it determines the amount of friction between the box and the ramp, which affects the speed at which the box slides down the ramp.

5. How is the acceleration of the box on a ramp calculated?

The acceleration of the box on a ramp can be calculated using the formula a = g sinθ - μcosθ, where a is the acceleration, g is the acceleration due to gravity, θ is the angle of the ramp, and μ is the coefficient of friction.

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