Magnitude of Force within incline?

[learningphysics]

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.

 

[johnsonandrew]

Fn- 54cos(34) - 20.3 = 0
Fn - 44.7 - 20.3= 0 <<----
Fn= 24.4

Edit: Nevermind I know what I did...

 

[anglum]

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

 

[johnsonandrew]

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

 

[learningphysics]

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...

 

[johnsonandrew]

Ohhh

 

[johnsonandrew]

So Fn - Fg parallel - Fg perpendicular = 0 ?

 

[anglum]

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

 

[johnsonandrew]

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