Find the kinetic friction coefficient

  • Thread starter Addez123
  • Start date
  • #1
106
11
Homework Statement:
See image1.
Block 1: 1M kg
Block 2: 2M kg
Block 3: 2M kg
Resulting acceleration: .5 m/s2
Relevant Equations:
F = ma
fk = Fn * u
1569828674457.png

Image1, the exercise.

My solution:
1569828729612.png

Image2, my solution

The answer is .37 but I keep getting .45
What am I doing wrong?
 

Answers and Replies

  • #2
collinsmark
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Homework Statement: See image1.
Block 1: 1M kg
Block 2: 2M kg
Block 3: 2M kg
Resulting acceleration: .5 m/s2
Homework Equations: F = ma
fk = Fn * u

View attachment 250394
Image1, the exercise.

My solution:
View attachment 250395
Image2, my solution

The answer is .37 but I keep getting .45
What am I doing wrong?
What's the total mass of the system?
 
  • #3
106
11
5M but I don't see your point?
I use F=ma on the center block (block 2), what does the total mass have to do with anything?
 
  • #4
haruspex
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what does the total mass have to do with anything?
All blocks will accelerate together, so their total mass is the total inertia.
Alternatively, assign unknowns to the tensions and draw a separate FBD for each block.
 
  • #5
106
11
It produce the correct answer, but I still don't see the corelation.
I mean we only look at forces applied to Block 2, but for some reason we need to consider the mass of the two other blocks aswell?

I mean F (from F=ma) here is applied only to block 2, its not applied to block 1 or 3. If it were they'd all move sideways. How come m should include block1 and 3 then?
 
  • #6
haruspex
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I mean we only look at forces applied to Block 2
But did you? The forces on block 2 are the tensions in the strings. They are not equal to the weights of the masses suspended on the ends.
As I wrote, draw separate FBDs for each block, treating the tensions as unknowns.
 

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