Block over another block, with friction force

[Emil233]

Homework Statement

An A block of mass 8 kg, is lying over an horizontal table, joint to a bucket of mass 1 kg, by a rope which passes by a pulley. The masses of the rope and the pulley are insignificant. Over the A block is lying a B block, of mass 2 kg. The coefficient of static friction, and kinetic friction, between the A and B blocks, and between the A block and the table, are static µ = 0.5, and kinetic µ = 0.4
1) If the bucket is loaded with 10 kg of sand, calculate the acceleration of the blocks and the bucket.
2) Show that for this load of sand, the B block does not slide over the A block.
3) What is the maximum load of sand that you can put in the bucket, without making the B block slide over the A block?

[PLAIN]http://a.imageshack.us/img825/7114/problemax.jpg

Homework Equations

F = ma, Frk = µk N, Frs max = µs N

The Attempt at a Solution

1) First, making a free body diagram for the blocks (where m blocks is the mass of the 2 blocks, and m bucket is the mass of the bucket loaded with sand(:

i) ΣFy = 0 = N - m blocks * g
N = m blocks * g

ii) ΣFx = T - Frk = m blocks * a (T = tension of the rope, Frk = kinetic friction force, between the table and the blocks)

And for the bucket:

ii) ΣFy = m bucket * g - T = m bucket a

So, T = -m bucket a + m bucket * g

Now, replacing T in ii)

-m bucket * g + mbucket * g - Frk = m blocks * a

-m bucket * g + mbucket * g - Frk - m blocks * a = 0

Frk is µk N, and N = m blocks * g, so:

-m bucket * a + m bucket * g - µk * m blocks * g - m blocks * a = 0

a (-m bucket - m blocks) + g (m bucket - µk * m blocks) = 0

a = (-g (m bucket - µk * m blocks)/(-m bucket - m blocks))

Now, m bucket is 11 kg, µk is 0.4, m blocks is 10 kg, and g = 9.81 m/s², so

a = (-9.81 m/s² (11 kg - 0.4 * 10 kg)/(-11 kg - 10kg)) = 3.27 m/s²

2) Now, making a free body diagram for the B block (m b block is the mass of the B block):

ΣFy = 0 = n - m B block * g
n = m B block * g

ΣFx = Fr s = m B block * a

m B block is 2 kg, and a = 3.27 m/s², so

Fr s = 2 kg * 3.27 m/s² = 6.54 N

Now, the maximum static friction force that can move the B block, is given by µs * n, where n is the normal force between the B block and the A block, and is equal to m B block * g.

So: Frs max = µs m B block * g = 0.5 * 2 kg * g = 9.81 N

The maximum static friction force, 9.81 N, is greater than the static force that moves the B block, which is 6.54 N. So, the B block does not slide over the A block.

3) Given that Frs = m B block * a, there is a maximum acceleration (a max) that the blocks can have, without making the B block slide over the A block.
So: Frs max = m B block * a max
µs * n = m B block * a max
µs * m B block * g = m B block * a max
µs * g = a max

Now, using this acceleration, and making a free body diagram for the bucket:

i) ΣFy = m bucket * g - T = m bucket * a max

And for the blocks:

ii) ΣFx = T - Frk = m blocks * a max
iii) ΣFy = N - m blocks * g = 0

T = m blocks * a max + Frk

Replacing T in i)

m bucket * g - m blocks * a max - Frk = m bucket * a max

m bucket * g - m blocks * a max - Frk - m bucket * a max = 0

m bucket * g - m blocks * a max - µk * m blocks * g - m bucket * a max = 0

m bucket (g - a max) - m blocks * a max - µk * m blocks * g = 0

m bucket = (+µk * m blocks * g + m blocks * a max) / (g - a max)

m bucket = (+0.4 * 10 kg * g + 10 kg * a max) / (g - a max)

m bucket = (+0.4 * 10 kg * g + 10 kg * µs * g) / (g - µs * g)

m bucket = (+0.4 * 10 kg * g + 10 kg * 0.5 * g) / (g - 0.5 * g) = 18 kg

Now, 18 kg is the maximum mass of the bucket allowed. The bucket alone has a mass of 1 kg, so, the maximum load of sand, would be 17 kg.

 

[collinsmark]

Hello Emil233,

Welcome to Physics Forums!

Your solution looks okay to me.