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## Homework Statement

You are given that the coefficient of kinetic friction between the block and the table in the attachment is 0.560, and m1 = 0.135 kg and m2 = 0.260 kg. (a) What should m3 be if the system is to move with a constant speed? (Assume ideal conditions for the string and pulleys.)

## Homework Equations

## The Attempt at a Solution

Start by writing down all of the forces involved in m1 and m2...

[itex]\Sigma F_{m1} = T_1 - m_1g = m_1a

[/itex]

[itex]

\Sigma F_{m2} = T_2 - m_2g = m_2a

[/itex]

[itex]

\Sigma F_{m3} = T_2 - T_1 - \mu m_3g = 0

[/itex]

I write = 0 because constant speed means no acceleration... so no acceleration means that F = ma => F = 0.

Now I have to solve this thingy, solve for T1 And T2 in the first two and stick it in the third.. getting..

[itex]

T_1 = m_1g + m_1a

[/itex]

[itex]

T_2 = m_2g + m_2a

[/itex]

[itex]

m_2g + m_2a - m_1g - m_1a - \mu m_3g = 0

[/itex]

Okie dokie, now we are able to solve for m3...

[itex]

m_3 = \frac{m_2g + m_2a - m_1g - m_1a}{\mu g} \rightarrow \frac{m_2g - m_1g}{\mu g}[/itex]

[itex]

\frac{1,225}{5.488} = 0.2232[/itex]