Question Video: Using Pascal's Principle to Determine the Force Needed to Lift an Object

What magnitude force must be exerted on the master cylinder of a hydraulic lift to support the weight of a car with a mass of 2500 kg resting on a second cylinder? The master cylinder’s diameter is 2.50 cm and the second cylinder’s diameter is 25.0 cm.

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Video Transcript

What magnitude force must be exerted on the master cylinder of a hydraulic lift to support the weight of a car with a mass of 2500 kilograms resting on a second cylinder? The master cylinder’s diameter is 2.50 centimeters and the second cylinder’s diameter is 25.0 centimeters.

From the information given, we can label the mass of the car 2500 kilograms as π‘š, the diameter of the master cylinder 2.50 centimeters as 𝑑 sub π‘š, and the diameter of the second cylinder 25.0 centimeters we’ll call 𝑑 sub two. We want to solve for the force magnitude that must be exerted on the master cylinder to accomplish this lift. We’ll call this force capital 𝐹 and start on our solution by drawing a diagram of the situation. In this scenario, we have a master cylinder where we apply a force and a secondary cylinder that does the lifting of the object we’re interested in elevating. In our case, that object is a car of mass π‘š of 2500 kilograms. We apply a force of magnitude 𝐹 to our master cylinder creating a pressure in our hydraulic system. That pressure is transmitted through this system and then acts to keep the car elevated.

To solve for the magnitude of that applied force 𝐹, we can recall that pressure is equal to force spread over an area 𝐴. And we also recall that in a hydraulic system such as we have here pressure is constant. That means if we choose any two points in our system, we’ll choose the input master cylinder and the output or secondary cylinder. The pressure at these two points, we can call them 𝑃 one and 𝑃 two, are the same. They’re equal to one another. This means that if we look at the area, where 𝑃 one is applied and we call that area 𝐴 sub π‘š, then we look at the area, where 𝑃 two is applied and we call that area 𝐴 sub two, by the equality of 𝑃 one and 𝑃 two, we can write 𝐹 over 𝐴 sub π‘š equals the force applied by the car, its mass times gravity, divided by 𝐴 sub two.

Rearranging this expression to solve for the force 𝐹, we see it’s equal to 𝐴 sub π‘š, the area of our master cylinder in cross section, divided by 𝐴 sub two, the cross-sectional area of our secondary lift, all multiplied by the weight force of the car π‘š times 𝑔. We can treat 𝑔, the constant, as exactly 9.8 meters per second squared. And to solve for 𝐴 sub π‘š and 𝐴 sub two, we remember that the cross sections of these cylinders are circles. We recall that the area of the circle is equal to πœ‹ times its diameter squared divided by four. The diameter of the master cylinder and the diameter of the secondary cylinder are given.

So we can rewrite our expression for force as πœ‹ over four 𝑑 sub π‘š squared over πœ‹ over four 𝑑 sub two squared all multiplied by π‘šπ‘”. And we see the factors of πœ‹ over four cancel out. Knowing 𝑑 sub π‘š, 𝑑 sub two, π‘š, and the acceleration due to gravity 𝑔, we’re ready to plug in and solve for 𝐹. When we do, entering these values on our calculator, we find that 𝐹 is 245 newtons. That’s the force we would need to apply to lift up a car of mass 2500 kilograms.

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