Finding the coefficient of friction

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  • #2
rock.freak667
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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?
 
  • #3
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Well its not at equilibrium so i can't just sum the forces to get zero ...?
 
  • #4
rock.freak667
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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'
 
  • #5
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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 :/
 
  • #6
rock.freak667
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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'.
 
  • #7
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Oh like net force?

The vertical would then be..

T - 0.3a = 0.3a ?
 
  • #8
rock.freak667
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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.
 
  • #9
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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
rock.freak667
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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
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You get

μ(2.94) + (2.94) = 0.6a ?
 
  • #12
rock.freak667
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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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