- #1
Karol
- 1,380
- 22
Homework Statement
m1 must stay in place. express m4 with the rest.
What must be the maximum ratio ##\frac{m_1}{m_2}## in order m1 will stay in place.
Homework Equations
Mass-acceleration: F=ma
The Attempt at a Solution
2 cases, in the first ##m_1<m_2##, which is drawn above:
$$T_1=m_1g,~~m_2g-T_1=m_2a_2~~\rightarrow~~a_2=\left( \frac{m_2-m_1}{m_2} \right)g$$
$$\frac{1}{2}a_2=a_3,~~2T_1=T_3,~~T_3-T_4=m_3a_3~~\rightarrow~~T_4=2T_1-\frac{1}{2}m_3a_3$$
$$T_4-m_4g=m_4a_3~~\rightarrow~~m_4=\frac{T_4}{a_3+g}=\frac{4m_1m_2-(m_2-m_1)m_3}{3m_2-m_1}$$
Since here ##m_1<m_2## the denominator of m4 is positive, so the nominator must be positive too:
$$4m_1m_2>(m_2-m_1)m_3,~~\frac{m_1}{m_2}\triangleq k~~\rightarrow~~k>\frac{m_3}{4m_2+m_3}$$
In the second scenario ##m_1>m_2##:
$$T_1-m_2g=m_2a_2~~\rightarrow~~a_2=\left( \frac{m_1-m_2}{m_2} \right)g$$
$$T_4-T_3=m_3a_3~~\rightarrow~~T_4=2T_1+\frac{1}{2}m_3a_3$$
$$m_4g-T_4=m_4a_3~~\rightarrow~~m_4=\frac{T_4}{g-a_3}=\frac{4m_1m_2+(m_1-m_2)m_3}{3m_2-m_1}$$
Since here ##m_1<m_2## the nominator is positive so the also must be the denominator, thus:
$$m_2-m_1~~\rightarrow~~\frac{m_1}{m_2}<3$$
Which doesn't make sense since the ratio doesn't include m3 or m4 at least, as in the previous case.