Acceleration of blocks on a slope

[Northbysouth]

Homework Statement

Two blocks are connected by a string that goes over an ideal pulley. Block A has a mass of 3.00 kg and can slide over a rough plane inclined at 30 degress to the horizontal. The coefficient of kinetic friction between block A and the plane is 0.400. block B has a mass of 2.77 kg. What is the acceleration of the blocks?

Homework Equations

f_k = u_kF_n

The Attempt at a Solution

F_n - 3gcos(30)
F_n = 25.4611

f_k = 0.4*25.4611
= 10.1845

Using force diagrams I identified the forces acting on the blocks to be:

T is the tension in the rope

For block A
T - f_k - 3gcos(30) = 3a

For block B
T - 2.77g = 2.77a

Then I combine them

f_k - 3gsin(30) -3a = 2.77g - 2.77a
a = 98.4

I know the answer is 0.392m/s^2 but I don't know where I'm going wrong. I assumed that block B is being pulled up because block A has a greater weight. Is this a wrong assumption?

 

[tiny-tim]

Hi Northbysouth!

f_k - 3gsin(30) -3a = 2.77g - 2.77a
a = 98.4

but you still need to find f_k

I assumed that block B is being pulled up because block A has a greater weight. Is this a wrong assumption?

depends on the slope …

obviously, if the slope was 0°, then B would win!

 

[Northbysouth]

I found f_k = 10.1845

Could you be more specific about the slope? How do I determine whether the friction is acting in the same direction as the tension or in the opposite direction?

 

[tiny-tim]

to find the direction of friction, imagine that there's no friction, and see which way the blocks go

(and I'm off to bed :zzz:)

 

[Northbysouth]

I ran the calculations again with friction acting opposite to the tension.

T - 10.1845 - 3gsin(30) =3a for block A

T =2.77a + 27.146 for block B

I substituted T into block A's equation:

2.77a+27.146 -10.1845 - 3gsin(30) = 3a
a = 0.520145

This still seems a ways off from 0.392. Am I missing something?

 

[tiny-tim]

(just got up :zzz:)

T - 10.1845 - 3gsin(30) =3a for block A

T =2.77a + 27.146 for block B

I substituted T into block A's equation:

2.77a+27.146 -10.1845 - 3gsin(30) = 3a
a = 0.520145

i] if you subtract equation B from A (so that the RHS is 3a), those first two terms on the LHS should be minus

ii] but your B equation is wrong if you're measuring a downwards (ie, if B is going down) …

then a = mg - T, not T - mg