Finding the Hill Incline Angle

In summary, the equation for power (P) is equal to the applied force (F) multiplied by the speed (V). In this problem, the car is experiencing a combined force of air resistance and friction while going up or down a hill at a constant speed of 27m/s. The car's engine needs to produce 47 hp more power going up the hill than it does going down, which is equivalent to 1298N of force. By setting up equations for the forces (gravity and friction) involved, it is determined that the angle of the hill above the horizontal is approximately 2.0 degrees.
  • #1
NotaPhysicsMan
146
0
This is supposed to be a hard question...

A 1900kg car experiences a combinded force of air resistance and friction that has the same magnitude whether the car goes up or down a hill at 27m/s. Going up a hill, the car's engine needs to produce 47 hp (1horsepower=745.7 Watts) more power to sustain the constant velocity than it does going down the same hill. At what angle is the hill inclined above the horizontal?

Ok, where do I even start?
 
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  • #2
The applied force times the speed is the power delivered by the vehicle. Simply write those out for both cases in terms of the gravitational force and the frictional force. Then simply subtract the equations! :-)
 
  • #3
but doesn't it have to be the average speed. P= F x Vave?
 
  • #4
Where does the 47 hp come into play?
 
  • #5
You said the speed is fixed and is the same for both the trip up and and the trip down the hill. You need to figure out what force the engine must exert for both cases and for each case the power delivered will be the respective force times that speed. Hint: Going up the hill the force is [itex]mg \cos \theta + F_{friction}[/itex]
 
  • #6
Hmm, are you sure it isn't mgsin(theta), the x-components right?. Going up hill, wouldn't friction oppose the x-force moving in another direction?
 
  • #7
Yes, of course it's sine and not cosine! Good catch!
 
  • #8
ok force of friction, I don't have the coeffecient of friction. Once I find the force to go uphill that's also equal to the force downhill with exception to the extra 47hp. Then I can separate into x and y compents. All I need is the force uphill or downhill.!
 
  • #9
you don't need to know anything about friction, since its force is the same in either direction and so will cancel. You don't know the total hp either, just the difference.
 
  • #10
ok I'm stuck here at this point: I have the F going down is equal to the one going up, right so let's see: (Engine force+1298N (the 47 hp)=the x component mgsin@. Here's where I'm stuck I don't know Engine force, and I don't konw @.
 
  • #11
ok according to Tide, The force going uphill is mgsin@+Ffriction

so the one going down hill is mgsin@-Ffriction

He said subtract the two, which doesn't make sense to me.
 
  • #12
I read the problem wrong again, let me fix it.

[tex] P_{uphill} - P_{downhill} = 47hp [/tex]

Yes read krab's post and the original problem to understand.
 
Last edited:
  • #13
Ok so let me see:

P=FV

P(uphill)=mgsin@ +F(friction)
P(downhill)=mgsin@-F(friction)

ok so I subtract them:

47hp=mgsin@+F - (mgsin@-F(friction)
47hp = 35047 Watts or 1298 N.

1298N=mgsin@-mgsin@+F+F
1298N=2F(friction)
F= 649 N. Funny Friction doesn't cancel out!

hmm,

Cos@=Wx/W
=649N/mg
=649N/18639N
=cos inverse(0.0348)
=88 degrees!

sounds kinda funny, that would be too steep.
 
  • #14
anyone want to verify?
 
  • #15
ah nuts, I think I did something wrong. Ok I think it should be sin@=Wx/W? that'll make it 2.0 degrees! humm. either too steep or too flat. What's wrong?
 
  • #16
NotaPhysicsMan said:
Ok so let me see:

P=FV

P(uphill)=mgsin@ +F(friction)
P(downhill)=mgsin@-F(friction)
What you are calling "P" is really the applied force, not the power. (Since you didn't multiply by V.) You need to rethink what force is needed in going up and down the hill.

Since the speed is constant, the applied force must just overcome friction and gravity. Going up, gravity acts against you (thus the applied force must be greater); going down, gravity acts in your favor (so the applied force is less). Friction always acts against you.
 
  • #17
ah dang I forgot about V! Ack I'm on it.
 
  • #18
k wait so all my other stuff are correct right? but I just didn't multiply by V? Adding V makes my angle pretty much 0!
 
  • #19
NotaPhysicsMan said:
k wait so all my other stuff are correct right? but I just didn't multiply by V? Adding V makes my angle pretty much 0!
Nope, not correct. You have the forces wrong. Reread my post, get the forces right, then worry about multiplying by V.
 
  • #20
P(uphill)=(F+1298)sin@ +F(friction)
P(downhill)=mgsin@-F(friction)

The only thing I see is that maybe I didn't add the force needed to get up there, the extra 47 hp or 1298N. Going downhill, I have friction against so that's neg.

a little help lol.
 
  • #21
hmm, well is my approach at least correct?
 
  • #22
I got it! the answer is 2.0 degrees?
Simple mistake really, didnt' take account of the 1298N and mixed up some of the negative signs.
 

1. What is the purpose of finding the hill incline angle?

The purpose of finding the hill incline angle is to determine the slope or steepness of a hill or incline. This information can be useful for various purposes such as construction, land surveying, or determining the difficulty level of a hiking trail.

2. How is the hill incline angle measured?

The hill incline angle can be measured using a clinometer, also known as an inclinometer, which is a device that measures angles of slope. Other methods include using a protractor and a string or using a smartphone app that has an inclinometer feature.

3. Why is it important to find the accurate hill incline angle?

Finding the accurate hill incline angle is important for safety reasons. Knowing the slope of a hill can help prevent accidents or injuries, especially for activities such as skiing, biking, or driving. It is also crucial for construction projects to ensure stability and proper drainage.

4. What factors can affect the hill incline angle?

The hill incline angle can be affected by various factors such as the height and length of the hill, the type of terrain, and the weather conditions. For example, a wet or icy surface can make a hill steeper and more difficult to climb.

5. Can the hill incline angle be calculated mathematically?

Yes, the hill incline angle can be calculated using basic trigonometry. The angle is equal to the inverse tangent of the rise (vertical distance) divided by the run (horizontal distance). However, this method may not be as accurate as using a specialized measuring tool.

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