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Finding the mininum value of the coffecitent of friction

  • Thread starter jtwitty
  • Start date
  • #1
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Homework Statement



1. Suppose a hanging 1.0kg lab mass is attached to a 4.0kg block on the table

(The picture)

lpZkr.jpg


a. If the coefficient of kinetic friction, [tex]\mu[/tex]k is .2, what is the acceleration?

b. What would the mininum value of the coefficient of static friction, [tex]\mu[/tex]s, in order for the block to remian motionless

Homework Equations



a = f/m
[tex]\mu[/tex] = frictional force / normal force

The Attempt at a Solution



a) .2 = friction/50 (i got 50 from 5 [the two blocks] multiplied by gravity)... friction = 10
a = 10/5
a = 2m/s

b) idk how to figure this out?
 

Answers and Replies

  • #2
You need to look at both blocks as a system.

Friction Force due to gravity on the hanging block.
<------------------(mass of both blocks)----------------------------------------------->

Keep in mind that a=f/m isn't really an equation. F=ma is the sum of all forces on an object or system. With that being said you can now use that equation as follows:

F=ma
Fg-Ff=ma....and I'm sure you can finish the rest.
 
  • #3
35
0
i meant

a = [tex]\Sigma[/tex]f / m

thats a real equation
 

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