Question

A pulley is used to lift a car engine . The engine is lifted 35.6 cm for every 1.07 m the rope is pulled . What is the IMA of the pulley

Answer 1

Gives

d₁=3.56m

d₂=1.07m

$W=F.dr$

W_in=W_out

$(F.d)_{in}=({F.d})_{out}$

Put value

$F_{in}\times3.56=F_{out}\times1.07$

$\frac{F_{out}}{F_{in}}=\frac{3.56}{1.07}=3.32$

We know that

IMA=$\frac{F_{out}}{F_{in}}$

Put value

$IMA=\frac{3.56}{1.07}=3.32$