Car going up a hill. Find the Angle of the hill.

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SUMMARY

The discussion focuses on calculating the angle of incline for a 1550-kg car traveling at a constant velocity of 21 m/s while experiencing forces due to air resistance and friction. The car's engine produces an additional 44 hp when ascending compared to descending the hill. The key equations used include P = FvCosθ and the components of weight force, specifically mgSinθ and mgCosθ. The final expression derived for power while going uphill is v(f + mgSinθ) = v(f - mgSinθ) + 32810.8 W, incorporating the power difference due to the incline.

PREREQUISITES
  • Understanding of Newton's Laws of Motion
  • Familiarity with power equations, specifically P = Fv
  • Knowledge of forces acting on inclined planes
  • Ability to manipulate trigonometric functions in physics contexts
NEXT STEPS
  • Study the principles of forces on inclined planes in physics
  • Learn about power calculations in mechanical systems
  • Explore the relationship between horsepower and watts for better conversions
  • Investigate the effects of friction and air resistance on vehicle dynamics
USEFUL FOR

This discussion is beneficial for physics students, automotive engineers, and anyone interested in understanding the dynamics of vehicles on inclines and the calculations involved in power and force analysis.

  • #31
No, Oh is it f=Fw+mgSin(theta)
 
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  • #32
So, Fw is?
 
  • #33
Fw=f-mgSin(theta)
 
  • #34
Good. So, find an expression for the power due to Fw going up and also for coming down. (Don't worry about numbers yet.)
 
  • #35
Up the hill: P=Fw(v) = (f+mgSinθ)v
Down the hill: P=Fw(v) = (f-mgSinθ)v
 
  • #36
Right. How do you now bring in the 44 hp?
 
  • #37
Set the equations equal to each other and add 44 to the "up hill" side
 
  • #38
Do you want to add it to the "up hill" side or the "down hill" side? Also, is hp the SI unit for power?
 
  • #39
o right becasue going uphill is 44 more hp you would add it to the downhill side. and no 1hp=745watt
 
  • #40
Right. (1 hp = 746 W the last I checked, but who's counting?) [OK, just checked, it's 745.699872 W for 1 "mechanical hp" and exactly 746 W for "electrical hp". Trivia for the day.]

http://en.wikipedia.org/wiki/Horsepower
 
Last edited:
  • #41
So just checking my equation:

v(f+mgSinθ)=v(f-mgSinθ) + 32810.8w

Now I can just plug and chug.
 
  • #42
Yep. Good work!
 
  • #43
I got a domain error... [never mind calculator error]
 
  • #44
What?
 
  • #45
I just put my nunmbers in my calculator wrong. No big deal thanks again.
 
  • #46
ok. No problem.
 

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