Question

Two blocks are placed on a smooth horizontal surface and connected by a string pulley arrangement as shown. If a force $F$ starts acting on block $m_1$, then find the relation between acceleration of both masses and their values.

Answer 1

From top to bottom, let us assume that the lengths of the strings are $l_1,l_2,l_3,l_4$

Therefore the total length is $l_1+l_2+l_3+l_4=l$.

After movement, lengths be $l_1-y,l_2-y+x,l_3-y+xand{l}_4-y+x$

So the total length will be $l_1-y+l_2-y+x+l_3-y+x+l_4-y+x$

Since $l_1+l_2+l_3+l_4=l$, we find that $y=3x/4$

Therefore if $a_1$ and $a_2$ are the accelerations, then $a_1$ = $\frac{d^{2}x}{d{t}^2}$ and $a_2=\frac{d^{2}y}{d{t}^2}$

So the relationship is $\frac{a_1}{a_2}=\frac{3}{4}$

Therefore, since acceleration of COM = $\frac{F}{(m1+m2)}$,

we can say that $a_1=\frac{F}{(m_1+(3m_2/4\left)\right)}$ and $a_2=\frac{3F}{(4m_1+3m_2)}$