Power Problem at inclined angle

In summary, a motorcycle with a mass of 232 kg and a rider is traveling at a steady speed of 21 m/s. The air resistance acting on the motorcycle and rider is 205 N. The power required to sustain this speed is 4305 W. However, if the road is sloped upward at 29.2° with respect to the horizontal, the power would need to be increased by finding the opposing force of the motorcycle's weight using W= F * cos29.2 = 232kg * 9.8 m/s^2 * cos29.2 * 21 m/s. This value would then need to be added to the original power calculated in part a. There is no need to consider
  • #1
emmi1987
4
0
A motorcycle (mass of cycle plus rider = 232 kg) is traveling at a steady speed of 21 m/s. The force of air resistance acting on the cycle and rider is 205 N. Find the power necessary to sustain this speed if (a) the road is level and (b) the road is sloped upward at 29.2° with respect to the horizontal.


I got through part a..
where P=Fv=205N * 21 m/s= 4305 W

but I'm not sure how to handle the opposing force of the motorcycle's weight down the slope...
I THINK you can find it with
W= F * cos29.2 s
= 232kg * 9.8 m/s^2 * cos29.2 * 21
(where 21=s given the velocity..ie distance per unit time?... or would it be 1/21 s in order to get rid of one of the 1/s)
THEN with this amount..
would you add it to part a?.. do you have to reconsider the air resistance since the bike is no longer traveling along the horizontal? (so.. 205N * sin29.2(?.. not cos because force is in opp direction) *21 m/s)..

anyway.. a bit confused.. any insight would be very appreciated I have to turn this in tonight. Thanks!
 
Physics news on Phys.org
  • #2
At first glance I believe you are correct in assuming W= F * cos29.2 s
= 232kg * 9.8 m/s^2 * cos29.2 * 21 m/s then adding this value to the original value calculated in part A. As for your question about units they do work out. Remember that a Newton is a kg m/s^2 so you end up with N*m/s or a j/s which is known as the watt!

Also there should be no need to take the sin component of the force due to air resistance because the air resistance would be a result of the movement of the motorcycle which means the force directly opposes the rider the same way that friction does... unless of course the problem is referring to a windy day.

edit: at second look taking the cos of 29.2 will give the wrong component of gravitational force that you are looking for
 
Last edited:
  • #3


I would approach this problem by breaking it down into its individual components and considering the forces acting on the motorcycle and rider.

First, let's consider the scenario where the road is level. In this case, the only force acting on the motorcycle is air resistance, which we have already calculated to be 205 N. Therefore, the power necessary to sustain this speed is simply P = Fv = 205 N * 21 m/s = 4305 W, as you have correctly calculated.

Now, let's consider the scenario where the road is sloped upward at 29.2° with respect to the horizontal. In this case, we need to take into account the force of gravity acting on the motorcycle and rider, which is equal to the weight of the system. As you have correctly calculated, the weight is given by W = 232 kg * 9.8 m/s^2 * cos29.2° = 1970 N.

Now, we need to consider the component of this weight that is acting in the direction of motion (along the slope). This can be found by multiplying the weight by the cosine of the angle of inclination, giving us Wcos29.2° = 1970 N * cos29.2° = 1756 N.

Therefore, the net force acting on the motorcycle and rider is the sum of the force of air resistance and the component of weight acting along the slope. This can be written as Fnet = 205 N + 1756 N = 1961 N.

Now, we can calculate the power needed to sustain this speed by using P = Fnetv = 1961 N * 21 m/s = 41181 W.

To summarize, when the road is sloped upward, the power necessary to sustain the speed of 21 m/s is 41181 W, which is significantly higher than the power required on a level road (4305 W). This is because the force of gravity is now acting against the motion, requiring more power to overcome it.

I hope this explanation helps you understand the problem better. I also recommend double-checking your calculations and considering the direction of the forces when solving problems like this. Good luck!
 

What is the "Power Problem at Inclined Angle"?

The "Power Problem at Inclined Angle" is a phenomenon in which the amount of power required to move an object up an inclined plane is greater than the power required to move the object on a horizontal surface. This is due to the effect of gravity on the object.

What causes the "Power Problem at Inclined Angle"?

The "Power Problem at Inclined Angle" is caused by the component of the object's weight that is acting against the direction of motion. This component, known as the "normal force," increases as the angle of the inclined plane increases, making it more difficult to move the object up the incline.

Why is it important to understand the "Power Problem at Inclined Angle"?

Understanding the "Power Problem at Inclined Angle" is important in many fields, including engineering, physics, and transportation. It allows for the calculation and optimization of power and energy requirements for moving objects on inclined surfaces, such as ramps and hills.

How can the "Power Problem at Inclined Angle" be overcome?

One way to overcome the "Power Problem at Inclined Angle" is by using mechanical advantage, such as a pulley system, to reduce the amount of force needed to move the object up the incline. Another solution is to decrease the angle of the inclined plane, reducing the effect of gravity on the object.

Are there any real-world applications of the "Power Problem at Inclined Angle"?

Yes, the "Power Problem at Inclined Angle" has many real-world applications. For example, it is important to understand when designing roads and highways to ensure that vehicles can safely navigate inclines. It also plays a role in the design of conveyor belts and other transportation systems.

Similar threads

Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
972
  • Introductory Physics Homework Help
Replies
4
Views
991
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
4K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
3K
  • General Math
Replies
4
Views
766
  • Introductory Physics Homework Help
Replies
8
Views
6K
Back
Top