A hydraulic jack is used to lift a car in which a man stands on one piston and car is put on another one. What should be the minimum weight of the man to just move the car upward?

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Solution

Given
$m_{1}$ = 800 kg
$A_{1}$ = 10 $m^{2}$
$A_{2}$ = 1 $m^{2}$ (1 Mark)

Using Pascak's Law,
$\frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}}$ .........(I) (0.5 Mark)

W know that F= m x a
$F_{1}$ = (800 X g) N
${F_{2}}_{2}$ = ( $m_{2}$ x g) N (0.5 Mark)

Put the value of $F_{1}$ and ${F_{2}}_{2}$ in (i)
$\frac{800 g}{10} = \frac{m_{2} g}{1}$

$\Rightarrow m_{2}$ = 80 kg (1 Mark)

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A hydraulic jack is used to lift a car in which a man stands on one piston and car is put on another one. What should be the minimum weight of the man to just move the car upward?

Given m1= 800 kg A1= 10 m2 A2 = 1 m2 (1 Mark) Using Pascak's Law, F1A1 = F2A2 .........(I) (0.5 Mark) W know that F= m x a F1 = (800 X g) N F2= (m2 x g) N (0.5 Mark) Put the value of F1 and F2in (i) 800 g10 = m2g1 ⇒m2 = 80 kg (1 Mark)