Question

An effort of 12.5kgf is applied on a machine through a distance of 60cm, when a load of 100kgf moves through a distance of 4m. Calculate: i) Mechanical advantage ii) Velocity ratio iii) Efficiency​

Answer 1

Answer:

Given :-

• An effort of 12.5kgf is applied on a machine through a distance of 60 cm, when a load of 100kgf moves through a distance of 4 m.

To Find :-

1. Mechanical Advantage
2. Velocity Ratio
3. Percentage efficiency of the machine

Solution :-

❶ Mechanical Advantage of the machine :-

As we know that :

$\clubsuit$ Mechanical Advantage or M.A Formula :

$\footnotesize\mapsto \sf\boxed{\bold{\pink{Mechanical\: Advantage =\: \dfrac{Load}{Effort}}}}$

Given :

• Load = 100kgf
• Effort = 12.5kgf

According to the question by using the formula we get,

$\implies \sf Mechanical\: Advantage =\: \dfrac{100}{12.5}$

$\implies \sf\bold{\red{Mechanical\: Advantage =\: 8}}$

${\small{\bold{\underline{\therefore\: The\: mechanical\: advantage\: of\: the\: machine\: is\: 8\: .}}}}$

❷ Velocity Ratio of the machine :

As we know that :

$\clubsuit$ Velocity Ratio or V.R Formula :

$\footnotesize\mapsto \sf\boxed{\bold{\pink{Velocity\: Ratio =\: \dfrac{Distance\: moved\: by\: effort}{Distance\: moved\: by\: load}}}}$

Given :

• Distance moved by effort = 60 cm
• Distance moved by load = 4 m

According to the question by using the formula we get,

$\longrightarrow \sf Velocity\: Ratio =\: \dfrac{\cancel{60}}{\cancel{4}}$

$\longrightarrow \sf Velocity\: Ratio =\: \dfrac{15}{1}$

$\longrightarrow \sf\bold{\red{Velocity\: Ratio =\: 15 : 1}}$

${\small{\bold{\underline{\therefore\: The\: velocity\: ratio\: of\: the\: machine\: is\: 15 : 1\: .}}}}$

❸ Percentage efficiency of the machine :

As we know that :

$\clubsuit$ Percentage Efficiency Formula :

$\footnotesize\mapsto \sf\boxed{\bold{\pink{Percentage\: Efficiency =\: \dfrac{Mechanical\: Advantage}{Velocity\: Ratio} \times 100}}}$

Given :

• Mechanical Advantage = 8
• Velocity Ratio = 15

According to the question by using the formula we get,

$\leadsto \sf Percentage\: Efficiency =\: \dfrac{8}{15} \times 100$

$\leadsto \sf Percentage\: Efficiency =\: \dfrac{800}{15}$

$\leadsto \sf\bold{\red{Percentage\: Efficiency =\: 53.33\%}}$

${\small{\bold{\underline{\therefore\: The\: percentage\: efficiency\: of\: the\: machine\: is\: 53.33\%\: .}}}}$

Answer 2

$\large\sf \underline \bold{ \underline{Given :}}$

Efforts = 8kgf

Loads = 100kgf

Distance through efforts = 50cm

Distance through loads = 3cm

 

$\large\sf \underline \bold{ \underline{To \: find :}}$

a)velocity ratio

b) mechanical advantage

c)percentage efficiency of the machine

   

$\large\sf \underline \bold{ \underline{solution :}}$

a) velocity ratio-

$\sf{Velocity \ radio = \dfrac{d_e}{d_l} }$

where, $d_e$ = Distance through effort moves

$d_l$= distance through load moves

$\sf{\implies V.R. = \dfrac{50}{3} }$

velocity ratio = 50:3

b) Mechanical advantage-

$\sf{M_{advantage} = \dfrac{load}{efforts} }$

$\sf{M_{advantage} = \dfrac{100}{8}= 12.5 }$

Mechanical advantage= 12.5

c)Percentage efficiency of the machine-

$\sf{ it \ is \ given \ by \rightarrow}$

$\sf{ \% = \dfrac {Mechanical \ advantage}{Velocity \ ratio} \times 100}$

$\sf{ \implies \dfrac{12.5}{\dfrac{50}{3}} \times 100}$

$\sf{\implies \dfrac{12.5}{16.66} \times 100}$

$\sf{\implies 75.03}$

Percentage efficiency of the machine is 75.03%