Question

2. The input and output pistons of a hydraulic jack are 4 cm and 12 cm in diameter. If a lever has a mechanical advantage of 9 and applied 200 Nof force on the input piston, how many pounds can the jack lift at 75% efficiency? 200 N of Force "MA" 9 12cm ? Newtons 4cm

Answer 1

Solution -

Input piston diameter (d) = 4 cm

Output piston diameter (D) = 12 cm

MA - 9

Input force (F) = 200 N

• Input force at the input piston (F') = input force x MA

= > 200 x 9 = 1800 N

• Input pressure = F' / a. ( a = area of input piston )
• a = π x (d/2)² =. 3.14 x (2 x 10^-2)²

= 12.56 x 10^-4 m^{​​​​​2}

• Input pressure will be -

= > 1800 / ( 12.56 x 10^-4 )

= 143.31 x 10⁴ N / m^{​​​​​​2}

^{According to Pascal's Law the same pressure will be transmitted to output piston ( Note here in calculation we are considering 100 % efficiency of the Jack , later at the end will incorporate the given efficiency that is 75 % )}

• Pressure at output piston ( P') - 143.31 x 10⁴ N/m^{​​​​​​2}​​​

• Load lifted = P' x A. ( A = Area of output Piston )

= >. A = π x (D/2)^2.

=. > 3.14 x ( 6 x 10^-2 )²

=. > 113. 04 x 10^-4 m^{​​​​​2}

• load lifted =  
• 113.04 x 10^-4 x 143.31 x 10⁴

^{​​​=. 16199.76 N}

^{The above answer came considering 100 % efficiency of Jack now if our jack is 75 % efficient then the load lifted would be multiplied with the efficiency factor of 0.75 to get the actual load lifted by the Jack .}

^{Actual load lifted = 16199.76 x 0.75}

^{=. 12149.82 N}

^{please appreciate efforts ? if you finds the solution conceptual and explained elicitly and solved your problem.}

^{​​​​​​​thank you}