Solving Friction Forces: Finding Coefficients for Block & Surface

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Homework Help Overview

The discussion revolves around a physics problem involving friction forces, specifically focusing on a 25.0-kg block on a horizontal surface. Participants are tasked with finding the coefficients of static and kinetic friction based on the forces required to initiate and maintain motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the setup of the problem using net force and equilibrium models. Questions arise regarding the expressions for static friction and the relationship between normal force and frictional force.

Discussion Status

The discussion includes attempts to clarify the expressions for static friction and the normal force. Some participants have provided guidance on how to relate these concepts, while others express uncertainty about the normal force and its role in the equations.

Contextual Notes

Participants are navigating through the problem setup and the definitions of forces involved, with some confusion about the normal force and its relationship to gravitational force. The discussion reflects a collaborative effort to clarify these concepts without reaching a definitive conclusion.

physics=world
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1. A 25.0-kg block is initially at rest on a horizontal surface.
A horizontal force of 75.0 N is required to set the block in
motion, after which a horizontal force of 60.0 N is required
to keep the block moving with constant speed. Find (a) the
coefficient of static friction and (b) the coefficient of
kinetic friction between the block and the surface.



2. Homework Equations



3. i need help setting up the problem. i would apply the net force model for the x direction and equilibrium model for the y model. i need to know if this is right.
Fx = 75.0N - F(static) - 60.0N = ma
Fy = n - mg = 0

 
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What is the expression for the coefficient of static friction?
 
is the expression for static friction just
-Fs = ma
 
physics=world said:
is the expression for static friction just
-Fs = ma
Not the above expression.
The expression for μ in terms of normal force and maximum frictional force.
 
rl.bhat said:
Not the above expression.
The expression for μ in terms of normal force and maximum frictional force.



oh okay. so i got

μ(s)n = ma

than i got n = mg for the second equation.

do i plug in "n" to the expression μ(s)n = ma ?

which would give me μ(s)mg = ma
 
physics=world said:
oh okay. so i got

μ(s)n = ma

than i got n = mg for the second equation.

do i plug in "n" to the expression μ(s)n = ma ?

which would give me μ(s)mg = ma
Correct. Now in the given problem, what is the normal force and ma?
 
rl.bhat said:
Correct. Now in the given problem, what is the normal force and ma?

hmm would mg = -245?
i don't know what the normal force is. i thought that mg and n always canceled to make it equal to zero.

opps. i meant would ma = 75
 
Normal reaction is equal and opposite to mg. mg acts on the floor and normal reaction acts on the block. Now you can find the coefficient of static friction.
 
rl.bhat said:
Normal reaction is opposite to mg. Now you can find the coefficient of static friction.

THANKS! i found it!:smile:
 

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