Finding the Accelation Of A Body In Presence Of Frictions

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Homework Help Overview

The discussion revolves around calculating the acceleration of a 60 kg block being pushed horizontally across a floor, considering the coefficients of static and kinetic friction. The original poster seeks to understand the correct approach to determine the acceleration, given the forces at play.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to calculate the total force needed by summing static and kinetic friction forces but questions the validity of their approach after arriving at an incorrect answer. Some participants clarify the roles of static and kinetic friction, suggesting that only kinetic friction is relevant once the block is in motion. Others question whether the constant force applied leads to a change in acceleration.

Discussion Status

Participants are exploring the implications of the initial force being just enough to overcome static friction and discussing the significance of using symbolic expressions in their calculations. There is recognition that some information may be superfluous, specifically regarding the mass of the block, which leads to a deeper understanding of the problem dynamics.

Contextual Notes

There are constraints regarding the use of units and the format of responses, as some participants have been reminded to avoid attaching images and to express their work symbolically. The discussion is also influenced by the need to clarify the relevance of different friction coefficients in the context of the problem.

Navin
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1. Homework Statement

A 60 kg block is pushed horizontally with just enough force to start its motion Accross a floor and the same force acts on it afterwards.The coefficient of static and sliding frictions are 0.5 and 0.4 respectively.
Find acceleration of the body.
Options

6m/s^2

4.9m/s^2

3.92m/s^2

1m/s^2

Homework Equations


F= ma

Friction = (u)m.a where " u " is coefficient of friction

The Attempt at a Solution


First I assumed that the total force need would be the summation of all the friction forces acting on the body

So I did
m.a = friction static + friction kinetic

60. a = 540

a = 9 m/s^2
But alas tis the wrong answer.

Then I tried another method which is in the attachment above I hope it is visible. ...but I'm pretty sure it's wrong.

[So don't delete the post oh ,moderator supreme overlord for I have tried !]

I would really like to know how to approach this problem

THIS IS ALL THE INFO GIVEN
 

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The static coefficient of friction gives you the maximal possible friction force when the object is stationary. The kinetic coefficient of friction gives you the friction when the object is moving. They will not both be relevant at the same time.
 
Orodruin said:
The static coefficient of friction gives you the maximal possible friction force when the object is stationary. The kinetic coefficient of friction gives you the friction when the object is moving. They will not both be relevant at the same time.
So when I make the equation for in motion I should just consider kinetic right ?

and secandly since the force is constant will there be change in accelaration to that of when it was set into motion ?
 
Think about how you will use each piece of information. (Although there is one piece of information that is superfluous, but let's find out which. The first statement is that the force is just sufficient to get the body moving in the first place. What does this tell you about the force with which you are pushing the body?
 
Orodruin said:
Think about how you will use each piece of information. (Although there is one piece of information that is superfluous, but let's find out which. The first statement is that the force is just sufficient to get the body moving in the first place. What does this tell you about the force with which you are pushing the body?
Well i think it tells us that the force is just sufficient to tip the balance.like its at the Citical value...therefore I think it's right to assume(please correct me if I'm wrong)
That the force will be equal to static friction
Hence
F=Friction (static)
F = 300
 
Navin said:
That the force will be equal to static friction
Yes, this is correct. The force is just sufficient to overcome static friction, which in effect is saying that it is essentially equal to the maximal static friction.

Navin said:
Hence
F=Friction (static)
F = 300
Units, please! Units are important when you do physics. If you do not use them, you will have a difficult time. Even better in this case would be to leave the symbolic expression for the force and only input numbers at the very end. This is often (very) helpful in understanding and finding errors. In this case, what is the symbolic expression for the force ##F## expressed in the mass ##m##, the gravitational acceleration ##g##, and the coefficient of static friction ##\mu_s##?
(If you are unfamiliar with LaTeX, you can use the symbols available when you press the ##\Sigma## in the editor to get μ and the ##x_2## symbol to make a subindex s, all in all giving μs.)
 
↔↔
Orodruin said:
Yes, this is correct. The force is just sufficient to overcome static friction, which in effect is saying that it is essentially equal to the maximal static friction.Units, please! Units are important when you do physics. If you do not use them, you will have a difficult time. Even better in this case would be to leave the symbolic expression for the force and only input numbers at the very end. This is often (very) helpful in understanding and finding errors. In this case, what is the symbolic expression for the force ##F## expressed in the mass ##m##, the gravitational acceleration ##g##, and the coefficient of static friction ##\mu_s##?
(If you are unfamiliar with LaTeX, you can use the symbols available when you press the ##\Sigma## in the editor to get μ and the ##x_2## symbol to make a subindex s, all in all giving μs.)
Great ! I think I got the answer (hopefully) the working is in the attachment

I'm sorry for not using units,I'll make sure to use them henceforth.im extremely unfamiliar with the keyboard layout !
 

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Please stop attaching pictures instead of typing your work. It violates forum policy.

Also, if you used symbolic expressions all the way rather than inserting numbers in every step, you would realize which piece of information in the problem was superfluous.
 
Orodruin said:
Please stop attaching pictures instead of typing your work. It violates forum policy.

Also, if you used symbolic expressions all the way rather than inserting numbers in every step, you would realize which piece of information in the problem was superfluous.
Gosh I should really read the terms and conditions ! Sorry for the attachments

mass is superfluous isn't it ?
 
  • #10
Navin said:
mass is superfluous isn't it ?
Yes. All of the forces are proportional to the mass, which means it cancels out with the ##m## in Newton's second law.
 
  • #11
Orodruin said:
Yes. All of the forces are proportional to the mass, which means it cancels out with the ##m## in Newton's second law.
Thank you so much ! This problem has been bugging me for a while .

Anyway have an awesome day !
 

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