# Magnitude of Force within incline?????

by anglum
Tags: force, incline, magnitude
 P: 71 53.9cos34 is greater than 1.45
 P: 275 im at a loss for # 2
 P: 88 20.3 !!! Check it...
 P: 275 20.3 is incorrect for the normal force otherwise known as #2
 P: 275 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
 P: 88 Damn I was sure that was it..
 P: 88 I said that 30.2^2 + F^2 = 36.4^2
 P: 275 if using what u listed johnson using the pythagorean theorem the Fn = 36.0987 is that what u get as well???
 P: 275 but that is also incorrect
 P: 88 Me? No I get 20.3 or 20.4
 P: 88 20.4 with correct significant figures... but I doubt it would be marked wrong for being only 0.1 off of the correct answer
 P: 88 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..
 P: 88 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.
 P: 275 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
 HW Helper P: 4,125 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?
 P: 88 dotted line is equal to parrelled force = Fg*sin x = 30.1964
 P: 88 F parallel is 30.1713, not 30.1964. AHh
 P: 88 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

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