Question

An effort of 8 KGF is applied on a machine through distance of 60cm when a load of 80 KGF moves through a distance of 4 cm. calculate

a. Velocity Ratio
b. mechanical advantage
c. efficiency the machine .​

Answer 1

$\underline{\large \sf { Solution- } }$

Given:-

• Efforts = 8kgf
• Loads = 100kgf
• Distance through efforts = 50cm
• Distance through loads = 3cm

To 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

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

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-

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

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

$\sf{ \pink\dashrightarrow \dfrac{12.5}{\dfrac{50}{3}} \times 100}$

$\sf{\pink\dashrightarrow 75.03}$

• Percentage efficiency of the machine is 75.03%

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

HOPES IT HELPS!:)

Answer 2

Answer:

Dil sa bulaya baalveer aaya