Question

An effort of 8kgf is applied ona machine through a distance of 50cm when load of 100kgf moves through a distance of 3cm . calculate a)velocity ratio . b) mechanical advantage c)percentage efficiency of the machine .​

Answer 1

Answer:

612 kg of the total weigh.

Explanation:

Answer 2

Given:-

☛Efforts = 8kgf

☛Loads = 100kgf

☛Distance through efforts = 50cm

☛Distance through loads = 3cm

 

Find:-

☛a)velocity ratio

☛b) mechanical advantage

☛c)percentage efficiency of the machine

   

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%