# Acceleration of blocks on a slope

1. Aug 9, 2012

### Northbysouth

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
fk = ukFn

3. The attempt at a solution

Fn - 3gcos(30)
Fn = 25.4611

fk = 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 - fk - 3gcos(30) = 3a

For block B
T - 2.77g = 2.77a

Then I combine them

fk - 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?

Last edited: Aug 9, 2012
2. Aug 9, 2012

### tiny-tim

Hi Northbysouth!
but you still need to find fk
depends on the slope …

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

3. Aug 9, 2012

### Northbysouth

I found fk = 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?

4. Aug 9, 2012

### 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:)

5. Aug 9, 2012

### 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?

6. Aug 10, 2012

### tiny-tim

(just got up :zzz:)
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