Finding the mininum value of the coffecitent of friction

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SUMMARY

The discussion focuses on calculating the minimum value of the coefficient of static friction (\mus) required for a 4.0kg block to remain motionless while a 1.0kg mass hangs from it. The coefficient of kinetic friction (\muk) is given as 0.2, leading to an acceleration calculation of 2 m/s² based on the frictional force derived from the total mass of the system. The participants emphasize the importance of treating both blocks as a single system to solve for static friction, using the equation F = ma to analyze forces acting on the blocks.

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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, \muk is .2, what is the acceleration?

b. What would the mininum value of the coefficient of static friction, \mus, in order for the block to remian motionless

Homework Equations



a = f/m
\mu = 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?
 
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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.
 
i meant

a = \Sigmaf / m

thats a real equation
 

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