Finding the coefficient of friction

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Homework Help Overview

The discussion revolves around finding the coefficient of friction in a scenario involving two blocks, one horizontal and one vertical, with forces acting on them. The original poster expresses confusion about how to approach the problem given the information provided, particularly focusing on the velocity aspect.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the forces acting on both blocks and the implications of non-equilibrium conditions. There are attempts to establish equations based on the forces and to relate them to acceleration. Questions arise about the relationships between tension, normal force, and friction.

Discussion Status

Participants are actively engaging with the problem, exploring different interpretations of the forces involved. Some guidance has been offered regarding setting up equations, but there is still uncertainty about how to eliminate variables and relate them to the coefficient of friction.

Contextual Notes

There is mention of specific values for forces and mass, but the original poster indicates a lack of clarity on how to proceed with the information given. The discussion reflects a need for further clarification on the relationships between the variables involved.

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Well on the horizontal block, what are the forces acting?

On the vertical block, what are the forces acting?

Find the resultant force on each block. What do you get?
 
Well its not at equilibrium so i can't just sum the forces to get zero ...?
 
xSnoopy said:
Well its not at equilibrium so i can't just sum the forces to get zero ...?

That is why you put the resultant force as 'ma'
 
so Force normal = 2.94N
and force moving horizontal would be (0.3)(a)

then
μ(2.94) + (0.3)(a) = (0.3)(a) ? ... crap I'm not getting something here :/
 
xSnoopy said:
so Force normal = 2.94N
and force moving horizontal would be (0.3)(a)

then
μ(2.94) + (0.3)(a) = (0.3)(a) ? ... crap I'm not getting something here :/

Consider the horizontal and vertical blocks separately.

On the horizontal block you have a tension T

so T-μ(0.3*9.81) = 0.3a

now do the same for the vertical block.

You will now have two equations in T and a. Eliminate 'T' from the two, leaving one equation in 'a'.
 
Oh like net force?

The vertical would then be..

T - 0.3a = 0.3a ?
 
xSnoopy said:
Oh like net force?

The vertical would then be..

T - 0.3a = 0.3a ?

Vertically, you'd get mg-T, since the block moves down.
 
So you would get

T-μ(0.3*9.81) = 0.3a
(0.3*9.81) - T = 0.3a

and substitute ?..

T-μ(0.3*9.81) = (0.3*9.81) - T

that doesn't cancel out the T's ?
 
  • #10
xSnoopy said:
So you would get

T-μ(0.3*9.81) = 0.3a
(0.3*9.81) - T = 0.3a

and substitute ?..

T-μ(0.3*9.81) = (0.3*9.81) - T

that doesn't cancel out the T's ?

No. If you add the two equations together, what do you get?
 
  • #11
You get

μ(2.94) + (2.94) = 0.6a ?
 
  • #12
right, so you can find 'a' in terms of 'μ'. You also know that 'a' is constant. Now use a kinematic equation involving distance, time and acceleration.

You should now be able to get the value of μ.
 

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