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Chapter 17: Problem 51

The drive belt on a vintage sander transmits \(1 / 2\) hp to a pulley that has a diameter of \(d=4\) in. Knowing that the pulley rotates at 1450 rpm, determine the tension difference \(T_{1}-T_{2}\) between the tight and slack sides of the belt.

Short Answer

Expert verified
The tension difference, \( T_1 - T_2 \), is approximately 48.43 N.

Step by step solution

01

Convert horsepower to watts

First, recognize that horsepower can be converted to watts since 1 horsepower equals 746 watts. Therefore, \( \frac{1}{2} \) hp can be converted as follows: \( 0.5 \times 746 = 373 \) watts.
02

Determine angular velocity in rad/s

The pulley rotates at 1450 revolutions per minute (rpm). Use the conversion factor \( \frac{2\pi}{60} \) to convert this to radians per second. Hence, \( \omega = 1450 \times \frac{2\pi}{60} \approx 151.65 \) rad/s.
03

Calculate the torque transmitted by the belt

The power transmitted by the belt is equal to torque times angular velocity: \( P = \tau \cdot \omega \). Rearrange to find torque, \( \tau = \frac{P}{\omega} = \frac{373}{151.65} \approx 2.46 \) N*m.
04

Relate torque to belt tension on the pulley

The torque on the pulley is also related by the equation \( \tau = (T_1 - T_2) \cdot r \), where \( r \) is the radius of the pulley. The radius \( r \) is half the diameter, i.e., \( \frac{4}{2} = 2 \) inches, which needs to be converted to meters as it equals \( 0.0508 \) meters. So, \( 2.46 = (T_1 - T_2) \times 0.0508 \).
05

Solve for tension difference

Rearrange the equation from Step 4 to solve for the difference in tension \( T_1 - T_2 \). Therefore, \( T_1 - T_2 = \frac{2.46}{0.0508} \approx 48.43 \) Newtons.

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Key Concepts

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

Power Transmission
Power transmission is a crucial concept in mechanics of materials. It refers to the movement of energy from one location to another, often to perform work. In our exercise, the drive belt on the sander transmits mechanical power efficiently. Power in this scenario is initially presented as horsepower, a common unit of power measurement in mechanical systems.
  • 1 horsepower equals 746 watts.
  • The belt transmits half a horsepower or 373 watts of power in this example.
Understanding how power is converted into the necessary work is essential. By knowing the power, you can determine how fast and how much work a system like the sander can do. In our case, converting power from horsepower to watts aids in subsequent calculations involving angular velocity and torque.
Angular Velocity
Angular velocity is a measure of how quickly an object rotates around a point. It is a key factor when dealing with rotating systems such as pulleys and wheels. The angular velocity is usually expressed in radians per second.
  • The vintage sander's pulley rotates at 1450 revolutions per minute (rpm).
  • To convert rpm to radians per second (rad/s), use the conversion factor: \( \omega = \text{rpm} \times \frac{2\pi}{60} \).
  • For our pulley: \( \omega = 1450 \times \frac{2\pi}{60} \approx 151.65 \) rad/s.
Understanding angular velocity helps in assessing how efficiently a system uses power. In this exercise, it provides a link between the pulley鈥檚 rotational speed and the torque exerted by the belt.
Torque Calculation
Torque is the force that causes an object to rotate around an axis. It's one of the primary factors used to determine the efficiency of a rotating system. In this exercise, torque is calculated using the power transmitted by the belt and the angular velocity.
  • The relationship between power, torque, and angular velocity is given by the equation \( P = \tau \cdot \omega \).
  • Rearranging gives \( \tau = \frac{P}{\omega} \).
  • Substituting our known values, \( \tau = \frac{373}{151.65} \approx 2.46 \) N*m.
This calculated torque represents the rotational force provided by the belt on the sander's pulley. Assessing torque is crucial in determining how much work the system will perform under load.
Belt Tension
Belt tension is critical in power transmission systems as it affects the effectiveness and safety of the belt-driven system. Tension is categorized into two sides:
  • **Tight Side (\( T_1 \))**: The side of the belt where tension is higher as it pulls the load.
  • **Slack Side (\( T_2 \))**: The side of the belt where tension is lower.
The difference in tension is essential to calculate the effectiveness of the power transmission. This is given by the equation involving torque: \( \tau = (T_1 - T_2) \cdot r \).
  • For our exercise, \( r \) is the radius of the pulley: 0.0508 meters.
  • Rearranging gives \( T_1 - T_2 = \frac{\tau}{r} \).
  • Substituting the torque value results in \( T_1 - T_2 = \frac{2.46}{0.0508} \approx 48.43 \) Newtons.
The calculation of belt tension difference ensures that the belt operates optimally without slipping or breaking, thus securing efficient power transmission.

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Most popular questions from this chapter

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