Newtons Law Involving Friction

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Homework Help Overview

The discussion revolves around a physics problem involving Newton's laws and friction, specifically focusing on a system of two blocks connected by a string over a frictionless pulley on an inclined plane. The original poster is attempting to calculate the minimum kinetic friction required to prevent the system from accelerating.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the problem and the expression for acceleration derived by the original poster. There are attempts to set the acceleration to zero to find the minimum kinetic friction coefficient. Some participants question the clarity of the setup and the algebraic steps taken.

Discussion Status

There is an ongoing exploration of the problem with various interpretations of the setup. Some participants have provided guidance on how to approach the calculations, while others express confusion regarding the algebra involved. The discussion includes attempts to reconcile discrepancies between personal calculations and textbook answers.

Contextual Notes

Participants note that the original poster's algebra may have led to incorrect results, and there is mention of specific values that do not match expected answers from a textbook or online platform. The problem includes conditions for acceleration direction based on the masses involved.

ISU20CpreE
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I have a block conected to another block by a string in an inclined surface, with a frictionless pulley. Both blocks are the same weight so i already figure the acceleration of the object that its being pulled up.

What I don't know is how can I calculate the minimum kinetic friction that will keep the system from accelerating.

I tried setting the acceleration to zero for this formula:

[tex]a= 1\ frac{m2*g-m1*g*sin\theta - \mu*m1*g*cos\theta} {m1+m2} [\tex]<br /> <br /> after that i really don't get the answer. Please any sugestions.[/tex]
 
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what no help??
 
Your description of the set up is a bit unclear, but if it's like what I have in the attachment, then I get the same expression as yourself for the acceleration.

If the system is not accelerating, then a = 0.
Set the numerator in your expression to zero and solve for [tex]\mu[/tex].
 

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Complete Problem

Fermat said:
Your description of the set up is a bit unclear, but if it's like what I have in the attachment, then I get the same expression as yourself for the acceleration.

If the system is not accelerating, then a = 0.
Set the numerator in your expression to zero and solve for [tex]\mu[/tex].

thats exactly what i did but it doesn't match the answer of the book.

this is the problem:

A block (mass m1) lying on a frictionless inclined plane is connected to a mass m2 by a massless cord passing over a pulley, as shown in Fig. 4-53.
(a) Determine a formula for the acceleration of the system in terms of m1, m2, , and g.
(b) What conditions apply to masses m1 and m2 for the acceleration to be in one direction (say, m1 down the plane), or in the opposite direction?

What smallest value of µk will keep the system of the previous problem from accelerating?
 
was your answer for the formula for acceleration, part (a), the same as in the book ?
 
if i try to see the smallest possible value of [tex]\mu[/tex] kinetic i get twp wrong answers "1.46" and "-0.61". It doesn't match my webassign page.
 
For [tex]\mu _k[/tex] I got

[tex]\mu = \left( \frac{m_2}{m_1}\right)sec\theta - tan\theta[/tex]

bit i can't see how I would get two answers ??
 
Fermet I got it. You were great help. I was wrong because my algebra procedure was wrong. I will be asking you more questions some of these days. For now I want to thank you for your help. Have a good one.

Best Regards,

Carlos
 
The answer is 0.63719 [tex]\approx[/tex] 0.64.
 
  • #10
ISU20CpreE said:
Fermet I got it. You were great help. I was wrong because my algebra procedure was wrong. I will be asking you more questions some of these days. For now I want to thank you for your help. Have a good one.

Best Regards,

Carlos
Ciao
scgyucfnbvs
 

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