Magnitude of Force within incline?

[johnsonandrew]

I'm working on it..

 

[anglum]

would it be sin34 = 30.1405/F to get #2?

thanks guys u are a great help... this is seriously the hardest class ever to take strictly online

 

[silvashadow]

same concept:
Fn=53.9cos34, does return a wrong answer from the website?

 

[johnsonandrew]

Whattt? I got 1.45.. I did something wrong

 

[silvashadow]

the funny part is that I am learning this right now, like 2 days ago

 

[johnsonandrew]

Haha yeah me too...that's why I'm doing it its good practice.

 

[silvashadow]

53.9cos34 is greater than 1.45

 

[anglum]

im at a loss for # 2

 

[johnsonandrew]

20.3 ! Check it...

 

[anglum]

20.3 is incorrect for the normal force otherwise known as #2

 

[anglum]

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

 

[johnsonandrew]

Damn I was sure that was it..

 

[johnsonandrew]

I said that 30.2^2 + F^2 = 36.4^2

 

[anglum]

if using what u listed johnson using the pythagorean theorem the Fn = 36.0987

is that what u get as well?

 

[anglum]

but that is also incorrect

 

[johnsonandrew]

Me? No I get 20.3 or 20.4

 

[johnsonandrew]

20.4 with correct significant figures... but I doubt it would be marked wrong for being only 0.1 off of the correct answer

 

[johnsonandrew]

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..

 

[johnsonandrew]

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.

 

[anglum]

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

 

[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