Magnitude of Force within incline?

In summary, the magnitude of the horizontal force F is 36.356N and the magnitude of the normal force Fn is 39.792N. These values were found using the sine and cosine functions, as well as the Pythagorean theorem, to solve for the forces acting on a block with a mass of 5.5kg on a frictionless incline with a slope of 34 degrees. However, initial incorrect answers were obtained due to using the incorrect angle of 56 degrees instead of 34 degrees.
  • #36
Haha yeah me too...that's why I'm doing it its good practice.
 
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  • #37
53.9cos34 is greater than 1.45
 
  • #38
im at a loss for # 2
 
  • #39
20.3 ! Check it...
 
  • #40
20.3 is incorrect for the normal force otherwise known as #2
 
  • #41
cant u use the pythagorean theorem to solve for # 2 since we know the horizontal is 36.356N

and the hypotenuse is what then? 53.9N

so woudlnt it be 53.9^2 = 36.356^2 +Fn^2

Fn = 39.792
 
  • #42
Damn I was sure that was it..
 
  • #43
I said that 30.2^2 + F^2 = 36.4^2
 
  • #44
if using what u listed johnson using the pythagorean theorem the Fn = 36.0987

is that what u get as well?
 
  • #45
but that is also incorrect
 
  • #46
Me? No I get 20.3 or 20.4
 
  • #47
20.4 with correct significant figures... but I doubt it would be marked wrong for being only 0.1 off of the correct answer
 
  • #48
But that has to be right: The line you get when you connect the horizontal force with the dotted line is the normal force. And it also forms a right triangle..
 
  • #49
If there is no penalty for entering wrong answers try 20.4 just in case... because I can't see how else to get the answer.
 
  • #50
i know i am totally puzzled on # 2

the hypotenuse we have established as 36.356N

the dotted line in the picture is equal to what?

the line which is the normal force squared is hyp squared minus dotted line squared
 
  • #51
did you get the first part? what did you get?

Suppose F is the horizontal force... what is the component of F perpendicular to the plane? What is the component of F parallel to the plane?
 
  • #52
dotted line is equal to parrelled force = Fg*sin x = 30.1964
 
  • #53
F parallel is 30.1713, not 30.1964. AHh
 
  • #54
F perpendicular is 54.0(cos 34), no? And wouldn't that be the normal force? If that's true we're making it more difficult than it is. But I guess we tried that earlier: 44.8 N
 
  • #55
learningphysics the answer to part 1 was 36.356N

so that is the hypotenuse in the triangle we have drawn

so we can square that and subtract the dotted line^2 from the picture on page 2 in this thread

and then solve for the Fn?
 
  • #56
johnsonandrew said:
Me? No I get 20.3 or 20.4

20.33 in other words Fsin(34) is the perpendicular component of F... there are 3 forces perpendicular to the plane... the normal force, mgcos(34) and 20.3...

Normal force - mgcos(34) - 20.3 = 0

solve for normal force
 
  • #57
Normal force then equals 24.4307N?
 
  • #58
anglum said:
Normal force then equals 24.4307N?

no. how did you get that?
 
  • #59
24.4 N?
 
  • #60
Fn -53.9cos(34) -20.3 = 0
Fn = 53.9cos(34) + 20.3
Fn = 44.685 + 20.3
Fn = 64.9851N
 
  • #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.
 
  • #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 [tex]\Sigma{Fy} = 0[/tex], 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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