Question

A small manufacturer is going to install a conveyer belt from their loading dock to the second floor to transport boxed components. When the components get to the second floor they will be unpacked, tested and assembled into the final product. The conveyer belt is 42 feet long and is supported on rollers every foot over the entire length. The conveyer belt is inclined 26.6 degrees. The boxes that will travel on the belt are each three feet long and weigh 62 pounds. When a shipment arrives, the conveyer belt at any instant in time will be jam filled until the entire shipment is on the second floor. The belt will often be stopped if the people on the second floor are not able to off-load the boxes fast enough. When they restart the belt, the belt will be fully loaded. Assume the coefficient of rolling resistance between belt and the rollers to be 0.02 and the belt full speed to be 0.5 feet per second. The acceleration time is 2 seconds. The drive roller at the end of the belt is 24 inches in diameter.

A. Calculate how much mechanical power in horsepower is necessary to accelerate the fully loaded belt to the desired final speed.

B. Calculate and draw the torque vs. rotational speed curve at the drive roller.

C. How much mechanical power in horsepower is necessary to keep the belt moving at full speed?

Answer 1