Magnitude of Force within incline?

[learningphysics]

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?

 

[johnsonandrew]

dotted line is equal to parrelled force = Fg*sin x = 30.1964

 

[johnsonandrew]

F parallel is 30.1713, not 30.1964. AHh

 

[johnsonandrew]

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

 

[anglum]

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?

 

[learningphysics]

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

 

[anglum]

Normal force then equals 24.4307N?

 

[learningphysics]

Normal force then equals 24.4307N?

no. how did you get that?

 

[johnsonandrew]

24.4 N?

 

[anglum]

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

 

[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..