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Magnitude of Force within incline? 
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#55
Sep2707, 10:08 PM

P: 275

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
Sep2707, 10:09 PM

HW Helper
P: 4,124

Normal force  mgcos(34)  20.3 = 0 solve for normal force 


#57
Sep2707, 10:10 PM

P: 275

Normal force then equals 24.4307N???



#58
Sep2707, 10:11 PM

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P: 4,124




#59
Sep2707, 10:13 PM

P: 88

24.4 N?



#60
Sep2707, 10:13 PM

P: 275

Fn 53.9cos(34) 20.3 = 0
Fn = 53.9cos(34) + 20.3 Fn = 44.685 + 20.3 Fn = 64.9851N 


#61
Sep2707, 10:14 PM

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#62
Sep2707, 10:15 PM

P: 88

Fn 54cos(34)  20.3 = 0
Fn  44.7  20.3= 0 << Fn= 24.4 Edit: Nevermind I know what I did... 


#63
Sep2707, 10:15 PM

P: 275

ok that is the correct answer
so our problem was we werent 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
Sep2707, 10:16 PM

P: 88

I'm so confused... I'm gonna go read over all that again haha..



#65
Sep2707, 10:18 PM

HW Helper
P: 4,124

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
Sep2707, 10:19 PM

P: 88

Ohhh



#67
Sep2707, 10:20 PM

P: 88

So Fn  Fg parallel  Fg perpendicular = 0 ?



#68
Sep2707, 10:20 PM

P: 275

thanks to all who have helped me .... tonite i finished earlier than usual... i got the other 9 problems sumwhat easily... thanks again



#69
Sep2707, 10:21 PM

P: 88

Thanks for posting it I learned something too. Thanks learningphysics..



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