Calculating Winch Power for an Inclined Ore Car: 30 Degree Incline | 950kg Mass

In summary: However, you did not use the correct formula for power, which is power = force * velocity. Also, your units were incorrect, as power should be in watts instead of kilowatts. In summary, the question asks for the power needed to move a loaded ore car up a 30.0 degree inclined mine shaft at a constant speed of 2.2 m/s. Using the formula power = force * velocity, we can find the force needed to pull the car, which is 4.9 N. Multiplying this by the velocity gives us a power of 10.78 W, which is slightly different from the given answer of 10.2 kW.
  • #1
cooltee13
25
0

Homework Statement


A loaded ore car has a mass of 950kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine shaft by a cable connected to a winch. The shaft is inclined at 30.0 degrees above the horizontal. The car accelerates uniformly to a speed of 2.2 m/s in 12.0 seconds and then continues at a constant speed.

a) what power must the winch motor provide when the car is moving at constant speed?


Homework Equations


Well, Power=Force x Velocity
Force = T

The Attempt at a Solution


Well to find the Tension in the rope, I found the component of G that was opposite the Tension and set it to -Fs. Then I set -Fs + T = 0 => Fs = T. Looking at The triangle under the cart I used sin(30)= Fs/g => Fs= gsin(30) which = 4.9 N

Since power = F*V I got P= (4.9)(2.2) = 10.78 kW. The back of the book says its 10.2 kW though. I am just wondering where I went wrong
 
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  • #2
cooltee13 said:
Well to find the Tension in the rope, I found the component of G that was opposite the Tension and set it to -Fs. Then I set -Fs + T = 0 => Fs = T. Looking at The triangle under the cart I used sin(30)= Fs/g => Fs= gsin(30) which = 4.9 N

I can't understand what you do with Fs and T. You do not use the mass of the cart. 4.9 N * 2.2 m/s will give you 10.78 W and not 10.78 kW
 
  • #3
Oh ok, well then care to give me some advice on how to solve the problem then?
 
  • #4
cooltee13 said:
Oh ok, well then care to give me some advice on how to solve the problem then?

The Idea of splitting the force of gravity in components parallel and perpendicular to the ground was ok. As was power = force * speed
 

1. What is the formula for calculating winch power for an inclined ore car?

The formula for calculating winch power for an inclined ore car is P = (m x g x sinθ x V) / η, where P is the winch power in watts, m is the mass of the ore car in kilograms, g is the acceleration due to gravity (9.8 m/s²), θ is the angle of incline in degrees, V is the velocity in meters per second, and η is the efficiency of the winch (usually between 0.8-0.9).

2. How do I determine the mass of the ore car?

The mass of the ore car can be determined by using a scale or by referring to the manufacturer's specifications. If the mass is not readily available, it can be estimated by multiplying the volume of the ore car by the density of the material it is carrying.

3. What is the significance of the incline angle in the equation?

The incline angle is a crucial factor in determining the winch power needed for an inclined ore car. The steeper the incline, the greater the force required to move the car, resulting in a higher winch power. This is because the force of gravity acting on the car increases as the angle of incline increases.

4. What is the role of winch efficiency in the equation?

Winch efficiency, denoted by η, is the ratio of the actual output power of the winch to the input power. In other words, it accounts for any energy losses in the winch system. A higher efficiency translates to a lower winch power requirement, as more of the input power is being used to move the ore car.

5. Can the formula be used for different incline angles and masses?

Yes, the formula can be used for any incline angle and mass, as long as the units are consistent. It is important to note that the velocity used in the formula should correspond to the velocity of the ore car at the incline, not the speed of the winch itself. Additionally, the efficiency value should be adjusted for the specific winch being used.

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