Question

A wooden block with mass 0.38 kg rests on a horizontal table, connected to a string that hangs vertically over a friction-less pulley on the table's edge. From the other end of the string hangs a 0.12 kg mass.

Find the block's acceleration if mu _k=0.20.

I did (m2*g - µ*m1*g )/(m2 - m1) = ( 0.12*9.8 - 0.20*0.38*9.8)/(0.38-0.12) = 1.658 m/s^2

but my homework online says I am wrong. What am I doing wrong?

Answer 1

Draw a FBD of both blocks.

Block that is hanging off the table:\[\sum F_y = m_1a = m_1g - T\]Block on table:\[\sum F_x = T - \mu m_2 g = m_2a\]

Assuming the wire between them is inextensible, both masses must have the same acceleration. Since the pulley is frictionless, the tension must be the same on both masses.

\[m_1(g-a) - \mu m_2g = m_2 a\]\[m_1g - \mu m_2 g = a(m_1+m_2)\]\[a = {m_1 g - \mu m_2 g \over m_1 + m_2}\]

Answer 2

Thanks a bunch I got the correct answer.

Answer 3

Great! Do you understand why I got \(m_1 + m_2\) in the denominator rather than \(m_2-m_1\)?

Answer 4

no not quite

Answer 5

Okay.

We know that the block that hangs from the pulley must pull the block on the table, because no external force acts on the block resting on the table. (Tension is not considered an external force.)

Therefore, if we let the positive y-direction be downwards for the hanging block, the acceleration of the block will be positive. Note that gravity acts downwards. Therefore\[ma = mg - T\]Tension is negative because it acts upwards.

Now, let's take the direction the tension acts to be positive for the block on the table. As established above, the block on the table will move in this direction. Therefore\[ma = T - \mu m g\]

To get the right sign of \(m_1 + m_2\), we must be careful that we are consistent in our direction of the acceleration of and the tension on the hanging block and the block on the table.

Answer 6

ohhh ok.. thanks for that explaination