The Radius Of The Effort Piston In A Hydraulic Lift Is Used To Raise A Load Of $120 \, \text{kg}$ With A Radius Of $2.5 \, \text{cm}$, Given That The Machine Is $80\%$ Efficient. Calculate:a) The Effort Needed.b) The Energy

Feb 26, 2025 by ADMIN 224 views

$The radius of the effort piston in a hydraulic lift is used to raise a load of $120 \, \text{kg}$ with a radius of $2.5 \, \text{cm}$, given that the machine is $80\%$ efficient. Calculate:a) The effort needed.b) The energy$

Introduction

Hydraulic lifts are widely used in various industries for lifting heavy loads with ease. The efficiency of a hydraulic lift is a crucial factor in determining the effort required to lift a load. In this article, we will calculate the effort needed to raise a load of $120 \, \text{kg}$ with a radius of $2.5 \, \text{cm}$ using a hydraulic lift that is $80\%$ efficient.

Theoretical Background

The effort required to lift a load using a hydraulic lift can be calculated using the principle of torque and rotational motion. The torque applied to the effort piston is equal to the torque required to lift the load. The torque required to lift the load is given by the product of the weight of the load and the radius of the load.

Formulae and Equations

The weight of the load is given by:

W = mg

where $m$ is the mass of the load and $g$ is the acceleration due to gravity.

The torque required to lift the load is given by:

\tau = Wr

where $r$ is the radius of the load.

The effort required to lift the load is given by:

F = \frac{\tau}{r_e}

where $r_e$ is the radius of the effort piston.

Calculation

Given that the load has a mass of $120 \, \text{kg}$ and a radius of $2.5 \, \text{cm}$ , we can calculate the weight of the load as follows:

W = 120 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 1176 \, \text{N}

The torque required to lift the load is given by:

\tau = 1176 \, \text{N} \times 0.025 \, \text{m} = 29.4 \, \text{Nm}

The effort required to lift the load is given by:

F = \frac{29.4 \, \text{Nm}}{0.025 \, \text{m}} = 1176 \, \text{N}

However, since the machine is $80\%$ efficient, the actual effort required is:

F_{actual} = \frac{1176 \, \text{N}}{0.8} = 1470 \, \text{N}

Energy Calculation

The energy required to lift the load is given by:

E = \frac{W \times h}{\eta}

where $h$ is the height through which the load is lifted and $\eta$ is the efficiency of the machine.

Assuming that the load is lifted through a height of $1 \, \text{m}$ , the energy required is:

E = \frac{1176 \, \text{N} \times 1 \, \text{m}}{0.8} = 1470 \, \text{J}

Conclusion

References

[1] "Hydraulic Lifts" by Wikipedia
[2] "Torque and Rotational Motion" by Physics Classroom
[3] "Efficiency of Machines" by HyperPhysics

Discussion

The calculation of the effort required to lift a load using a hydraulic lift is a complex process that involves the principles of torque and rotational motion. The efficiency of the machine is a crucial factor in determining the actual effort required. In this article, we have calculated the effort needed to raise a load of $120 \, \text{kg}$ with a radius of $2.5 \, \text{cm}$ using a hydraulic lift that is $80\%$ efficient.

Limitations

The calculation of the effort required to lift a load using a hydraulic lift assumes that the machine is $80\%$ efficient. However, in reality, the efficiency of the machine may vary depending on various factors such as the design of the machine, the quality of the components, and the operating conditions.

Future Work

Future work in this area may involve the development of more efficient hydraulic lifts that can reduce the effort required to lift heavy loads. This may involve the use of advanced materials and designs that can improve the efficiency of the machine.

Conclusion

In conclusion, the effort needed to raise a load of $120 \, \text{kg}$ with a radius of $2.5 \, \text{cm}$ using a hydraulic lift that is $80\%$ efficient is $1470 \, \text{N}$ . The energy required to lift the load is $1470 \, \text{J}$ . The calculation of the effort required to lift a load using a hydraulic lift is a complex process that involves the principles of torque and rotational motion. The efficiency of the machine is a crucial factor in determining the actual effort required.

Q&A

Q: What is the weight of the load?

A: The weight of the load is given by the product of its mass and the acceleration due to gravity. In this case, the mass of the load is $120 \, \text{kg}$ and the acceleration due to gravity is $9.8 \, \text{m/s}^2$ . Therefore, the weight of the load is:

W = 120 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 1176 \, \text{N}

Q: What is the torque required to lift the load?

A: The torque required to lift the load is given by the product of the weight of the load and the radius of the load. In this case, the weight of the load is $1176 \, \text{N}$ and the radius of the load is $0.025 \, \text{m}$ . Therefore, the torque required to lift the load is:

\tau = 1176 \, \text{N} \times 0.025 \, \text{m} = 29.4 \, \text{Nm}

Q: What is the effort required to lift the load?

A: The effort required to lift the load is given by the product of the torque required to lift the load and the radius of the effort piston. In this case, the torque required to lift the load is $29.4 \, \text{Nm}$ and the radius of the effort piston is $0.025 \, \text{m}$ . Therefore, the effort required to lift the load is:

F = \frac{29.4 \, \text{Nm}}{0.025 \, \text{m}} = 1176 \, \text{N}

However, since the machine is $80\%$ efficient, the actual effort required is:

F_{actual} = \frac{1176 \, \text{N}}{0.8} = 1470 \, \text{N}

Q: What is the energy required to lift the load?

A: The energy required to lift the load is given by the product of the weight of the load, the height through which the load is lifted, and the efficiency of the machine. In this case, the weight of the load is $1176 \, \text{N}$ , the height through which the load is lifted is $1 \, \text{m}$ , and the efficiency of the machine is $80\%$ . Therefore, the energy required to lift the load is:

E = \frac{1176 \, \text{N} \times 1 \, \text{m}}{0.8} = 1470 \, \text{J}

Q: What are the limitations of this calculation?

A: The calculation of the effort required to lift a load using a hydraulic lift assumes that the machine is $80\%$ efficient. However, in reality, the efficiency of the machine may vary depending on various factors such as the design of the machine, the quality of the components, and the operating conditions.

Q: What are the future directions of this research?

A: Future work in this area may involve the development of more efficient hydraulic lifts that can reduce the effort required to lift heavy loads. This may involve the use of advanced materials and designs that can improve the efficiency of the machine.