/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 The driveshaft of a building's a... [FREE SOLUTION] | 91Ó°ÊÓ

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The driveshaft of a building's air-handling fan is turned at 300 RPM by a belt running on a \(0.3\)-m-diameter pulley. The net force applied by the belt on the pulley is \(2000 \mathrm{~N}\). Determine the torque applied by the belt on the pulley, in \(\mathrm{N} \cdot \mathrm{m}\), and the power transmitted, in \(\mathrm{kW}\).

Short Answer

Expert verified
The torque applied by the belt is 300 N·m, and the power transmitted is 9.426 kW.

Step by step solution

01

- Calculate the radius of the pulley

The radius of the pulley can be calculated by dividing the diameter by 2. Given the diameter is 0.3 meters, the radius is:\( R = \frac{0.3}{2} = 0.15 \) meters.
02

- Calculate the torque applied by the belt on the pulley

Torque (\( \tau \)) is the product of the force applied and the radius of the pulley. The force applied by the belt is 2000 N and the radius is 0.15 meters:\( \tau = \text{Force} \times \text{Radius} = 2000 \times 0.15 = 300 \) N·m.
03

- Convert RPM to radians per second

First, convert the rotational speed from revolutions per minute (RPM) to radians per second. Given 300 RPM:\( \text{Angular velocity} = 300 \times \frac{2\pi}{60} = 300 \times \frac{\pi}{30} \approx 31.42 \) radians per second.
04

- Calculate the power transmitted by the belt

Power (P) is the product of torque (\( \tau \)) and angular velocity (\( \omega \)). We've calculated the torque as 300 N·m and angular velocity as 31.42 radians per second:\( P = \tau \times \omega = 300 \times 31.42 \approx 9426 \) W or 9.426 kW.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Torque Calculation
Torque is a measure of the rotational force applied to an object. It can be calculated using the formula: \( \tau = F \times r \).
Here, \( \tau \) is the torque, \( F \) is the force applied, and \( r \) is the radius. In this exercise, the pulley has a radius of 0.15 meters and the force applied by the belt is 2000 N. Thus, the torque is calculated as:
\( \tau = 2000 \times 0.15 = 300 \) N·m.
Torque is essential in understanding how rotational systems behave under different forces, and it determines the ability to cause an object to turn about an axis.
Pulley Systems
Pulley systems are mechanical devices used to change the direction of force and to increase mechanical advantage in lifting or moving objects. A simple pulley consists of a wheel with a groove for the belt or rope that carries the load.
In the given problem, a pulley with a 0.3-meter diameter is used. The net force applied by the belt is 2000 N. The pulley's radius is important for calculating torque. The radius is half of the diameter: \( r = \frac{d}{2} = \frac{0.3}{2} = 0.15 \) meters.
Pulley systems simplify the effort needed in mechanical power transmission, allowing for efficient and effective movement of equipment in various applications.
Rotational Dynamics
Rotational dynamics involves the study of forces and torques and their effect on rotational motion. Key quantities are angular velocity, torque, and moment of inertia. In this problem, the drive shaft is rotated at 300 RPM by the belt.
Converting this speed to radians per second is essential for further calculations in rotational dynamics. The conversion formula is:
\( \omega = RPM \times \frac{2\pi}{60} \).
For 300 RPM, the angular velocity \( \omega \) is:
\( \omega = 300 \times \frac{2\pi}{60} = 31.42 \) radians per second.
This angular velocity helps in calculating the power transmitted by the belt in our rotating system, demonstrating the interconnectedness of rotational dynamics concepts.
Power Transmission
Power transmission in mechanical systems involves transferring energy from one component to another through forces and motion. Power (P) in rotational systems is calculated using torque and angular velocity:
\( P = \tau \times \omega \).
In this exercise, the torque applied by the belt is 300 N·m and the angular velocity is 31.42 radians per second. Thus, the power transmitted by the belt is:
\( P = 300 \times 31.42 = 9426 \) W, or 9.426 kW.
Efficient power transmission ensures the effective functioning of mechanical systems, reducing energy loss and mechanical wear, and is crucial for the smooth operation of machinery.
Angular Velocity
Angular velocity (ω) describes the rate of rotation of an object and is measured in radians per second. It is converted from revolutions per minute (RPM) using the formula:
\( \omega = RPM \times \frac{2\pi}{60} \).
For instance, a rotational speed of 300 RPM is converted to:
\( \omega = 300 \times \frac{2\pi}{60} = 31.42 \) radians per second.
Angular velocity links the rotational speed to other rotational dynamics quantities, such as torque and power, and helps in assessing the performance of rotating machinery.

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Most popular questions from this chapter

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