Question

You are given that the coefficient of kinetic friction between the block and the table in Fig. 4.36 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?
(b) If m3 = 0.095 kg, what is the magnitude of the acceleration of the system? (Assume ideal conditions for the string and pulleys.)

Answer 1

i'm not sure
a. 0.129
b. 1.46

Answer 2

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

\[\Sigma F_{m1} = T_1 - m_1g = m_1a\]
\[\Sigma F_{m2} = T_2 - m_2g = m_2a\]
\[\Sigma F_{m3} = T_2 - T_1 - \mu m_3g = 0\] 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..

\[T_1 = m_1g + m_1a\]
\[ T_2 = m_2g + m_2a\]

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

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

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

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