NEIEP Mechanics Practice Exam

If a machine's drive sheave has a circumference of six feet, how many RPMs are needed to run the elevator car at 800 feet per minute?

1. 100 RPMs

2. 133 RPMs

3. 150 RPMs

4. 200 RPMs

The correct answer is: 133 RPMs

To determine the required RPM (revolutions per minute) to achieve an elevator car speed of 800 feet per minute with a drive sheave circumference of six feet, you can use the formula to connect linear speed and rotational speed. First, recall that the linear speed of a point on the circumference of the sheave when it rotates is equal to the circumference multiplied by the number of revolutions per minute (RPM). The relationship can be expressed as: Linear Speed (feet per minute) = Circumference (feet) × RPM By rearranging this formula to find RPM, we have: RPM = Linear Speed / Circumference Substituting the known values into the formula: - Linear Speed = 800 feet per minute - Circumference = 6 feet This gives us: RPM = 800 feet per minute / 6 feet RPM = 133.33 Since RPM is typically expressed as a whole number, you would round down to 133. Therefore, running the drive sheave at approximately 133 RPM will allow the elevator car to move at its desired speed of 800 feet per minute. This calculation confirms that the answer of 133 RPM is indeed accurate, illustrating the relationship between linear and