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Help with finding the coefficient of kinetic friction!

  • #1
ScreenShot2012-12-11at104406AM.png


Ok so I tried making myself a free-body diagram but I reached a dead end..

Step 1) fkinetic = μkn
fkinetic = μk(251)
fkinetic / 251 = μk

Step 2)
Fx:
T1

Fy
n = (m)(g) = 251 N
120.5 N

Step 3)

ƩFx = m(-a) since it's moving the left or ƩFx = 0 (since it's moving at CONSTANT velocity)
ƩFy = ma

Step 4)

ƩFy: 251 + 120.5 = (25.58 + 12.28)a

*** I got the masses by solving 120.5 = m(9.81) and 251 = m(9.81) ***

ƩFy: 371.5 = (37.86)a
ƩFy: 371.5/37.86 = a
ƩFy: 9.81 = a


Step 5)

T1 = m(-a)
T1 = (25.58)(-9.81) = 250.9 = 251 :(

Help?
 

Answers and Replies

  • #2
265
0
One thing I would advise you to do is not to substitute the numbers in until the end - it's better to work with symbols as long as possible.

Using your free-body diagrams, write out equations for the horizontal and vertical forces acting on W1 and the vertical forces acting on W2.
Then see how you can combine them to get a simple expression for μ
 
  • #3
One thing I would advise you to do is not to substitute the numbers in until the end - it's better to work with symbols as long as possible.

Using your free-body diagrams, write out equations for the horizontal and vertical forces acting on W1 and the vertical forces acting on W2.
Then see how you can combine them to get a simple expression for μ
W1

Forces in x direction
T1
fk

Forces in y direction
Fn
_________________________

W2

Forces in x direction
none

Forces in y direction
T2
_________________________

ƩFx: T1 + fk = ?
T1 + 251μk = ?​
ƩFy: Fn + T2 = ?
(m1g) + (-m2a) = ?​
(25.58*9.81) + (-12.28*9.81) = ?​
251 -120.5 = ?​
130.5 = ?​


What next?
 
Last edited:
  • #4
265
0
ƩF = ma. In this case, you're told that a=0, and this applies to both masses.
Also, you need to remember the direction of the forces.
So,
ƩFx = T1 - fk = 0
→ T1 = fk

Similar problem with signs for the forces on W2
 
  • #5
ƩF = ma. In this case, you're told that a=0, and this applies to both masses.
Also, you need to remember the direction of the forces.
So,
ƩFx = T1 - fk = 0
→ T1 = fk

Similar problem with signs for the forces on W2
Ohh ok.
 
  • #6
ƩF = ma. In this case, you're told that a=0, and this applies to both masses.
Also, you need to remember the direction of the forces.
So,
ƩFx = T1 - fk = 0
→ T1 = fk

Similar problem with signs for the forces on W2
Ohh ok.


ƩFx: T1 + (-fk) = ma
T1 - 251μk = m(0)​
T1 - 251μk = 0​
T1 = 251μk

Wait but I still have to find T1...How do I do that?

ƩFy: Fn + T2 = ?
(m1g) + (-m2a) = ?​
(25.58*9.81) + (-12.28*9.81) = ?​
251 - 120.5 = ?​
130.5 = ?​
 
  • #7
265
0
Your first equation is ok.

Apply the same reasoning to the second equation:
ƩFy = T2 - W2 = 0
→ T2 = W2

Now, T1 = T2 since the pulley just changes the direction of the tension force and not its value.

So, you can combine the 2 equations to eliminate the tension force - the answer should be clear after doing that.
 
  • #8
Your first equation is ok.

Apply the same reasoning to the second equation:
ƩFy = T2 - W2 = 0
→ T2 = W2

Now, T1 = T2 since the pulley just changes the direction of the tension force and not its value.

So, you can combine the 2 equations to eliminate the tension force - the answer should be clear after doing that.
T1 = T2
T1 = 120.5

So for

T1 = 251μk
120.5 = 251μk
0.48 = μk

Thanks!

One quick question though...
Whenever I see a problem like this - two blocks and a pulley...

T1 always equals T2?
 
  • #9
265
0
As long as the rope and pulley are 'ideal', ie don't have any mass, then the 2 tensions will be the same.
 
  • #10
As long as the rope and pulley are 'ideal', ie don't have any mass, then the 2 tensions will be the same.
Ok thanks a bunch!!
 

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