Power Developed for Conveyor Belt Transporting Newspapers

  • Thread starter Lothar
  • Start date
In summary, the conveyor belt has a length of 13.5 m and is inclined at 30.0°, transporting bundles of newspapers with a mass of 1.0 kg each and 26 newspapers per bundle. The task is to determine the power developed by the conveyor if it delivers 15 bundles per minute. Using the equation P = w/t, where w is the work and t is the time, the work can be calculated by finding the change in potential energy of one newspaper when moving to the top of the ramp. This is given by W = m*g*Δh, where m is the mass of the newspaper, g is the acceleration due to gravity, and Δh is the change in height of the
  • #1
Lothar
19
0

Homework Statement


A 13.5 m long conveyor belt, inclined at 30.0°, is used to transport bundles of newspapers from the mailroom up to the cargo bay to be loaded on to delivery trucks. Each newspaper has a mass of 1.0 kg, and there are 26 newspapers per bundle. Determine the power that the conveyor develops if it delivers 15 bundles per minute.

Homework Equations


w = fd
w = fdcos(theta)
P = w/t

The Attempt at a Solution


I've tried several things getting incorrect answers of:
57.33
859.95
744.74
49.6
372.4
 
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  • #2
Lothar said:

Homework Statement


A 13.5 m long conveyor belt, inclined at 30.0°, is used to transport bundles of newspapers from the mailroom up to the cargo bay to be loaded on to delivery trucks. Each newspaper has a mass of 1.0 kg, and there are 26 newspapers per bundle. Determine the power that the conveyor develops if it delivers 15 bundles per minute.

Homework Equations


w = fd
w = fdcos(theta)
P = w/t

The Attempt at a Solution


I've tried several things getting incorrect answers of:
57.33
859.95
744.74
49.6
372.4

Welcome to PF.

Maybe focus on your last equation? If you figure the work performed in 1 minute you're just about home free right?

So what is the work of moving 1 newspaper to the top of the ramp?

W = ΔPE = m*g*Δh

where Δh = 13.8*sin30 = 1/2*13.8 = 7.9 m

Figure then the total work from 15*26 newspapers and that tells you how much for a minute, so getting to work/sec is really easy then.
 
  • #3


I would approach this problem by first identifying the known variables and equations that can be used to solve for the power developed by the conveyor belt. The known variables are the length of the conveyor belt (13.5 m), the angle of inclination (30.0°), the mass of each newspaper (1.0 kg), and the number of newspapers per bundle (26). The unknown variable is the power developed by the conveyor belt.

Based on the given information, we can use the equation P = w/t, where P is power, w is work, and t is time. In this case, we need to calculate the work done by the conveyor belt in a given time (t), which is equal to the force (f) applied multiplied by the distance (d) traveled.

First, we need to calculate the force (f) applied by the conveyor belt to move the newspapers. This can be done using the equation w = fd, where w is the weight of the newspapers. The weight of one newspaper is equal to its mass (1.0 kg) multiplied by the acceleration due to gravity (9.8 m/s^2). Therefore, the weight of one newspaper is 9.8 N. The total weight of 26 newspapers is 9.8 N x 26 = 254.8 N.

Next, we need to calculate the distance (d) traveled by the conveyor belt in one minute. Since the conveyor belt is inclined at 30.0°, we can use the equation w = fdcos(theta) to calculate the horizontal distance traveled. The angle of inclination (30.0°) is the same as the angle between the horizontal and the hypotenuse of the triangle formed by the conveyor belt. Therefore, the horizontal distance traveled in one minute is d = 13.5 m x cos(30.0°) = 11.68 m.

Now, we can plug in the values for weight (254.8 N) and distance (11.68 m) into the equation P = w/t and solve for power. The time (t) is given as 1 minute, or 60 seconds. Therefore, P = (254.8 N x 11.68 m)/60 s = 49.93 watts.

The power developed by the conveyor belt is 49.93 watts. This means that the conveyor belt is capable of delivering 49.
 

1. How is power developed for conveyor belt transporting newspapers?

The power for conveyor belt transporting newspapers is typically developed through the use of electric motors. These motors turn the rollers that move the conveyor belt, allowing for the continuous and efficient transport of newspapers.

2. What type of power source is used for conveyor belt transporting newspapers?

As mentioned earlier, electric motors are the most common power source used for conveyor belt transporting newspapers. However, some smaller conveyor belts may also be powered by air or hydraulic systems.

3. How much power is needed for conveyor belt transporting newspapers?

The amount of power needed for conveyor belt transporting newspapers depends on several factors such as the length and weight of the conveyor belt, the speed at which it needs to operate, and the incline or decline of the conveyor. Generally, electric motors used for this purpose range from 1-10 horsepower.

4. Can renewable energy sources be used to power conveyor belts transporting newspapers?

Yes, renewable energy sources such as solar panels or wind turbines can be used to power conveyor belts transporting newspapers. However, the feasibility of using these sources depends on the location and availability of the renewable energy source.

5. How is the power consumption monitored for conveyor belt transporting newspapers?

Power consumption for conveyor belt transporting newspapers can be monitored through the use of energy meters, which measure the amount of electricity used by the conveyor belt. This data can then be analyzed to identify ways to increase efficiency and reduce energy costs.

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