Magnitude of Force within incline?

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SUMMARY

The discussion centers on calculating the magnitude of the horizontal force (F) and the normal force (Fn) acting on a 5.5 kg block on a frictionless incline of 34 degrees. The correct calculations involve using the sine and cosine of the incline angle: Fh = Fg * sin(34) = 30.2 N and Fn = Fg * cos(34) = 44.8 N. Participants emphasized the importance of using the correct angle in calculations and accounting for all forces acting perpendicular to the incline to achieve accurate results.

PREREQUISITES
  • Understanding of basic physics concepts such as force, weight, and equilibrium.
  • Familiarity with trigonometric functions (sine and cosine) and their application in physics.
  • Knowledge of free-body diagrams and how to analyze forces acting on an object.
  • Ability to perform calculations involving gravitational force (Fg = mass * gravity).
NEXT STEPS
  • Study the application of Newton's laws of motion in inclined planes.
  • Learn about free-body diagrams and how to represent forces visually.
  • Explore the concept of static equilibrium and how it applies to forces on inclines.
  • Practice solving problems involving inclined planes with varying angles and friction.
USEFUL FOR

Students in physics courses, educators teaching mechanics, and anyone seeking to understand forces on inclined planes and their calculations.

  • #61
anglum said:
Fn -53.9cos(34) -20.3 = 0
Fn = 53.9cos(34) + 20.3
Fn = 44.685 + 20.3
Fn = 64.9851N

yes. that's correct.
 
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  • #62
Fn- 54cos(34) - 20.3 = 0
Fn - 44.7 - 20.3= 0 <<----
Fn= 24.4

Edit: Nevermind I know what I did...
 
Last edited:
  • #63
ok that is the correct answer

so our problem was we weren't calculating all 3 perpendicular forces... we were just finding one of them and using that as Fn...

however its all 3 of them equal to 0
 
  • #64
I'm so confused... I'm going to go read over all that again haha..
 
  • #65
anglum said:
ok that is the correct answer

so our problem was we weren't calculating all 3 perpendicular forces... we were just finding one of them and using that as Fn...

however its all 3 of them equal to 0

Yes, it is important to go with the basic equations, and get the results from there... don't take shortcuts...

the most important step is \Sigma{Fy} = 0, where y is perpendicular to the plane... plug in your forces into the left side... Fn - mgcos(34) - Fsin(34) = 0...
 
  • #66
Ohhh
 
  • #67
So Fn - Fg parallel - Fg perpendicular = 0 ?
 
  • #68
thanks to all who have helped me ... tonite i finished earlier than usual... i got the other 9 problems sumwhat easily... thanks again
 
  • #69
Thanks for posting it I learned something too. Thanks learningphysics..
 

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