Magnitude of Force within incline?????

silvashadow

the funny part is that im 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?