Physics, asked by suryajai285, 5 hours ago

A hydraulic system is used to lift a 2000 kg vehicle in an auto garage. If the vehicle sits on a piston of area 0.5 m2 , and a force is applied to a piston of area 0.03 m2 , what is the minimum force that must be applied to lift the vehicle? Given: Area covered by the vehicle on the piston A1 = 0.5 m2, Weight of the vehicle, F1 = 2000 kg × 9.8 m s–2;

Answers

Answered by Afreenakbar

Answer:

The minimum force that must be applied to lift the vehicle using the hydraulic system is approximately 1176 Newtons.

Explanation:

Pascal's law, which states that pressure applied to an enclosed fluid is transferred equally in all directions, can be used to determine the minimal force necessary to lift the car using the hydraulic system.

The following provides the pressure formula:

$Pressure = \frac{Force }{ Area}$

The pressure placed on the larger piston in the hydraulic system is equivalent to the pressure placed on the smaller piston. We can construct the equation shown below:

Pressure on the larger piston = Pressure on the smaller piston

$\frac{ Force \: on \: the \: larger \: piston }{ Area \: of \: the \: larger \: piston }= \frac{Force \: on \: the \: smaller \: piston }{ Area \: of \: the \: smaller \: piston}$

Let's write F₂ for the force acting on the larger piston, A₂ for the area of the larger piston, and A1 for the force acting on the smaller piston.

The formula for the vehicle's weight is F₁ = m × g, where m is the vehicle's mass and g is the acceleration brought on by gravity.

Given:

Area of piston 1 (A₁) = 0.5 m²

Weight of the vehicle (F₁) = 2000 kg × 9.8 m/s²

Area of piston 2 (A₂) = 0.03 m²

Now, we can determine the bare minimum force needed:

$F_2 = ( \frac{ F_1 }{ A_1}) \times A_2$

$F_2 = ( \frac{2000 kg \times 9.8 m/s^2 }{ 0.5 m^2}) \times 0.03 m^2$

F2 ≈ 1176 N

As a result, 1176 Newtons or less of force is required to elevate the vehicle utilising the hydraulic system.