Friction Formulas: Fn, Ff & Mu w/ D, T, A, Vi & Ramp Angle

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The discussion focuses on deriving the formulas for normal force (Fn), friction force (Ff), and the coefficient of friction (mu) for an object sliding down a ramp, given specific variables including distance (D), time (T), acceleration (A), initial velocity (Vi), and ramp angle. Participants clarify the definitions of the variables and share attempts at formulating the equations, noting challenges with incorrect results. Key equations involve trigonometric functions to resolve forces acting on the object, specifically the components of weight parallel and perpendicular to the ramp surface. The importance of ensuring that the coefficient of friction remains less than one is emphasized, along with the suggestion to avoid complicating the problem by separating forces into x and y components. The discussion aims to refine the approach to accurately calculate the forces involved in the scenario.
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Homework Statement


What is the formula for: Fn, Ff, and mu when you have a object sliding down a ramp.
Variable know are D, T, A, Vi, and the angle of the ramp above the horrizon.


The Attempt at a Solution


I've tried a few that ended up being quite long and all wrong...
 
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Lobber said:

Homework Statement


What is the formula for: Fn, Ff, and mu when you have a object sliding down a ramp.
Variable know are D, T, A, Vi, and the angle of the ramp above the horrizon.

What are D, T, A? I take it Fn is the normal force, Ff is the friction force and Vi is the initial velocity?

Lobber said:

The Attempt at a Solution


I've tried a few that ended up being quite long and all wrong...

Please could you show some of your working?

You should obtain expressions involving trig. functions and the properties of your mass.
 
D=1.83M
T=1.06S
A=3.257m/s/s
angle/theta is 54 degrees
Vi=0
M=0.5kg
Friction force?
Normal force?
Mu?

Formula#1 (didn't work...)
Ff:
(cos theta A)*M=Max=Fx
Fx=Ffx-Fgx
Fgx=(sin theta)Mg.
Fn:
Ma(sin theta) = May
May=Fy=Fny-Fgy
Fgy=M*(cos theta)g

Ff/Fn=mu

Formula2

Fx=Max
ax=a(cos theta)
Fy=May
ay=a(sin theta)
Fg=Mg
Fx=Max=Fgx-Ffx
Fgx=Mg(sin theta)
Fy=May=Fgy-Fny
Fgy=Mg(cos theta)
Ff/Fn=mu.

My answers with these two formulas were ff=4.93 Fn=4.20 mu=1.20

for formula 2 they were Ff=3 Fn=1.56 mu=1.92
 
I think I might have fixed my problem as ay=Fgy=0 so that changes things. Also it's an inclined plane question.
 
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Your value of mu should always be less than one.

I think you need to find the difference between the actual force at the bottom and the theoretical force at the bottom.

Please could you post equations that would give the component of weight normal to the surface and the component of wieght parallel to the surface and we will go from there.

Edit: don't split it up into x and y parts, it makes things more messy.
 
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Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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