Help with finding the coefficient of kinetic friction

[riseofphoenix]

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

Step 1) f_kinetic = μ_kn
f_kinetic = μ_k(251)
f_kinetic / 251 = μ_k

Step 2)
F_x:
T₁

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

Step 3)

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

Step 4)

ƩF_y: 251 + 120.5 = (25.58 + 12.28)a

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

ƩF_y: 371.5 = (37.86)a
ƩF_y: 371.5/37.86 = a
ƩF_y: 9.81 = aStep 5)

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

Help?

 

[ap123]

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

 

[riseofphoenix]

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
f_k

Forces in y direction
F_n
_________________________

W₂

Forces in x direction
none

Forces in y direction
T₂
_________________________

ƩF_x: T₁ + f_k = ?

T₁ + 251μ_k = ?​

ƩF_y: F_n + T₂ = ?

(m₁g) + (-m₂a) = ?​

(25.58*9.81) + (-12.28*9.81) = ?​

251 -120.5 = ?​

130.5 = ?​

What next?

 

[ap123]

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

 

[riseofphoenix]

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.

 

[riseofphoenix]

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.ƩF_x: T₁ + (-f_k) = ma

T₁ - 251μ_k = m(0)​

T₁ - 251μ_k = 0​

T₁ = 251μ_k​

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

ƩF_y: F_n + T₂ = ?

(m₁g) + (-m₂a) = ?​

(25.58*9.81) + (-12.28*9.81) = ?​

251 - 120.5 = ?​

130.5 = ?​

 

[ap123]

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.

 

[riseofphoenix]

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 = 251μ_k
120.5 = 251μ_k
0.48 = μ_k

Thanks!

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

T1 always equals T2?

 

[ap123]

As long as the rope and pulley are 'ideal', ie don't have any mass, then the 2 tensions will be the same.

 

[riseofphoenix]

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!