Easy Acceleration / Friction question

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Homework Help Overview

The discussion revolves around a physics problem involving a box on an inclined roof, specifically addressing the effects of friction and acceleration. The original poster presents calculations related to gravitational force, normal force, and acceleration without friction, seeking guidance on incorporating friction into their analysis.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of Newton's second law and the role of friction in determining acceleration. The original poster expresses uncertainty about how to incorporate the coefficient of friction into their calculations.

Discussion Status

Some participants provide hints and guidance on using Newton's second law and calculating frictional force. The discussion is ongoing, with multiple interpretations of how to approach the problem being explored.

Contextual Notes

The original poster has made several calculations but is unsure about the next steps, particularly regarding the friction coefficient. There is an emphasis on understanding the relationship between forces and acceleration in the context of the problem.

dglenn9000
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Homework Statement



A roof has an incline of 10 degrees. If the µ of friction between a 50kg box and roof is .15, how fast will the box move?

Homework Equations



Ff=µ * Fn

The Attempt at a Solution



I have no idea how to get this. I know it very easy.
I figured out Fg = 490N (Fg=9.8 * 50kg)
I figured out Fn = 482.56 (Fn=490 cos 10)
I figured out F//=85.09 (F//=490 sin 10)

I figured out the Acceleration with no friction is 1.70 m/s2 85.09/50

Any help is appreciated, thanks
 
Last edited:
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Welcome to PF!

Hi dglenn9000! Welcome to PF! :wink:
dglenn9000 said:
A roof has an incline of 10 degrees. If the µ of friction between a 50kg box and roof is .15, how fast will the box move?

Hint: just use good ol' Newton's second law …

total force (in the direction of the slope) = mass times acceleration :smile:
 
I have a questions : there is an air track and on that track there is a body at rest upon which acts a constant force for a given time it gains a velocity. My question is if that force acts again upon that body for that amount of time will the velocity be 2 times the first velocity? My opinion is that starting from Newtons law v=v0+at, but if we use two times this quation we obtain that v1=at+at=2v0. Am I right?
 
Welcome to PF!

Hi DIrtyPio! Welcome to PF! :smile:

(but please start a new thread in future)
DIrtyPio said:
… a constant force for a given time … if that force acts again upon that body for that amount of time will the velocity be 2 times the first velocity? My opinion is that starting from Newtons law v=v0+at, but if we use two times this quation we obtain that v1=at+at=2v0. Am I right?

yes, but it's better to look at it from the words of Newton's second law …

force = rate of change of momentum = mass times rate of change of velocity,

so (from the meaning of "rate of change") ∫force times time = mass times change of velocity. :wink:
 


tiny-tim said:
Hi dglenn9000! Welcome to PF! :wink:


Hint: just use good ol' Newton's second law …

total force (in the direction of the slope) = mass times acceleration :smile:


That is how I got how fast it will move without friction.

a= F/m
1.70 m/s2= 85.09/50kg

But my question is how do I figure out the acceleration with friction of .15? Where do I apply the .15 in my equation?
 
Hi dglenn9000! :smile:

just woke up :zzz: …
dglenn9000 said:
… Where do I apply the .15 in my equation?

You multiply it by the normal force to get the (magnitude of the) friction force: Ffriction = µN.

You then chuck the friction force into the "total force" in Ftotal = ma :wink:
 
Hi everyone do you do any problems considering phyisc?
 

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