Grade 8 Hydraulics

A hydraulic lift works by applying force to one end of a closed tube to create a stronger force on the other end. In order to lift a car, a hydraulic lift must exert a force equal to the mass of the car. If a car has a weight of 1,375N what force must be applied to the other end if the area of the end lifting the car is 10 m² and the area of the other end is 2 m²?

• 1100N

• 825N

• 550N

• 275N

Answer explanation

To find the force needed, use the formula: Force = (Area of other end / Area of lifting end) * Weight of the car. Substituting the values, we get Force = (2 / 10) * 1375 = 275N. Therefore, the correct answer is 275N.

A tube has an area of 1.0 m² on one end and an area of 0.25 m² on the other. In order for a force of 27N to be created on the 0.25 m² end, what force would the force have to be on the 1.0 m² end?

• 108N

• 36N

• 81N

• 54N

Answer explanation

To create a force of 27N on the 0.25 m^2 end, the force on the 1.0 m^2 end would be 4 times that amount, which is 108N. Therefore, the correct choice is 108N.

A piston in a large engine moves with a force of 500N on an area of 5 m². To balance this force, what force must be applied on the other side of the piston with an area of 2 m²?

• 200N

• 1000N

• 250N

• 125N

Answer explanation

To balance the force, the pressure on both sides of the piston must be equal. Using the formula Pressure = Force / Area, the force on the other side with an area of 2 m^2 is (500N * 5 m^2) / 2 m^2 = 1000N.

If a hydraulic system is designed to amplify force, and a user applies a force of 50N on one end with an area of 0.5 m², what is the force exerted on the other end with an area of 2 m²?

• 200N

• 100N

• 400N

• 50N

Answer explanation

The force exerted on the other end is calculated by multiplying the force applied by the user (50N) by the ratio of the areas (2m^2 / 0.5m^2), resulting in 200N.

In a hydraulic brake system, a force of 100N is applied to a piston of area 1 m². What is the force generated on a piston at the other end with an area of 0.2 m²?

• 20N

• 500N

• 200N

• 50N

Answer explanation

The force on the second piston can be calculated using the formula F2 = F1 * (A2 / A1), which gives F2 = 100N * (0.2m^2 / 1m^2) = 20N. Therefore, the correct answer is 200N.

When a force of 300N is applied to a piston with an area of 3 m², what force is needed on a piston with an area of 1.5 m² to maintain equilibrium?

• 150N

• 600N

• 450N

• 300N

Answer explanation

To maintain equilibrium, the pressure on both pistons must be equal. Using the formula F1/A1 = F2/A2, we find that 300N/3m^2 = F2/1.5m^2, giving F2 = 600N.

If a hydraulic press has a small piston of area 0.5 m² and a large piston of area 2.5 m², what force must be applied to the small piston to lift a 1000N weight placed on the large piston?

• 200N

• 100N

• 50N

• 400N

Answer explanation

To lift the 1000N weight, the pressure on both pistons must be equal. Using the formula F1/A1 = F2/A2, we find that F1 = (1000N * 2.5m^2) / 0.5m^2 = 200N.

In a hydraulic system, if a force of 200N is applied to a piston of area 2 m², what is the resulting force on a piston at the other end with an area of 0.5 m²?

• 50N

• 800N

• 400N

• 200N

Answer explanation

The pressure is constant in a hydraulic system, so using the formula F1/A1 = F2/A2, we find F2 = (200N * 2m^2) / 0.5m^2 = 800N. Therefore, the resulting force on the other piston is 800N.

Given a hydraulic system where a force of 120N is applied to a piston of area 1.5 m², what force is exerted on a piston at the other end with an area of 3 m²?

• 60N

• 240N

• 180N

• 120N

Answer explanation

The force exerted on the other piston can be calculated using the formula: Force = Pressure x Area. Therefore, the force exerted on the other piston is 60N (120N/2).

If a force of 400N is applied to a piston with an area of 4 m² in a hydraulic system, what is the force on a piston at the other end with an area of 2 m²?

• 200N

• 800N

• 400N

• 600N

Answer explanation

The force is transmitted equally in a hydraulic system, so the force on the other piston is 800N (400N * 4m^2 / 2m^2).