Hydraulic Piston Supporting a 500lb weight

In summary, the conversation discusses a problem involving two pistons with different diameters and a weight that needs to be supported. The equation F1/A1=F2/A2 is used to determine the force needed, but the initial attempt yields an incorrect answer. The conversation then explores the idea of using torque to solve the problem, but this also results in an incorrect answer. Ultimately, it is determined that the ratio of force on the piston to the force at the hand is equal to the ratio of their distances from the pivot.
  • #1
AsuraSky
16
0

Homework Statement



Piston 1 in the figure below has a diameter of 0.31 in.; Piston 2 has a diameter of 1.7 in. In the absence of friction, determine the force vector F necessary to support the 500 lb weight.

p9_24.gif


Homework Equations



F1/A1=F2/A2



The Attempt at a Solution



I used the equation above and found the areas of each piston and solved for F1


F1/.075477=500/2.2698

F1=16.626

That answer was wrong so I thought about it some more and realized that the force I solved for was the force on the piston and not the lever that the question asked for. So I had this crazy idea that since there was sort of a rotational force on the hinge when you push the lever down you can use Torque = rF with torque being the force previously solved for

16.626 = 10(F) (r=10 for the 10 in distance between force F in the diagram and piston 1 in the diagram)

F= 1.6626

This answer was also marked wrong and I have run out of ideas on how to solve this problem. If anyone can point me in the right direction it would be greatly appreciated.
 

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  • #2
Right method, slight oops.

The ratio of force at the piston and the hand is the ratio of the distances from each of them to the pivot.
 
  • #3
Thanks, that cleared things up for me.
 

1. How does a hydraulic piston support a 500lb weight?

A hydraulic piston supports a 500lb weight through the use of hydraulic pressure. The piston is connected to a hydraulic system, which utilizes a liquid, usually oil, to transmit force. When a force is applied to one end of the piston, it creates pressure in the hydraulic fluid, causing the piston to extend and support the weight.

2. What is the maximum weight that a hydraulic piston can support?

The maximum weight that a hydraulic piston can support depends on the strength and size of the piston, as well as the hydraulic system it is connected to. Typically, hydraulic pistons can support weights ranging from a few pounds to several thousand pounds.

3. Are there any limitations to using a hydraulic piston to support heavy weights?

One limitation of using a hydraulic piston to support heavy weights is the potential for leaks in the hydraulic system. If there is a leak in the system, it can cause a loss of pressure and weaken the support of the weight. Additionally, the weight must be evenly distributed on the piston to prevent strain and potential failure.

4. What are the benefits of using a hydraulic piston to support a 500lb weight?

Using a hydraulic piston to support a 500lb weight offers several benefits. Hydraulic systems are highly efficient and can transmit a lot of force with minimal energy input. This makes them ideal for supporting heavy weights. Additionally, hydraulic pistons are compact and can be easily controlled and adjusted, making them a versatile choice for various applications.

5. How do I maintain a hydraulic piston supporting a 500lb weight?

To maintain a hydraulic piston supporting a 500lb weight, it is important to regularly check the hydraulic fluid levels and ensure there are no leaks in the system. It is also important to evenly distribute the weight on the piston and avoid overloading it. Regular maintenance and inspections can help prolong the lifespan of the hydraulic piston and ensure safe operation.

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