Magnitude of Force within incline?????


by anglum
Tags: force, incline, magnitude
silvashadow
silvashadow is offline
#37
Sep27-07, 09:53 PM
P: 71
53.9cos34 is greater than 1.45
anglum
anglum is offline
#38
Sep27-07, 09:54 PM
P: 275
im at a loss for # 2
johnsonandrew
johnsonandrew is offline
#39
Sep27-07, 09:54 PM
P: 88
20.3 !!! Check it...
anglum
anglum is offline
#40
Sep27-07, 09:56 PM
P: 275
20.3 is incorrect for the normal force otherwise known as #2
anglum
anglum is offline
#41
Sep27-07, 09:57 PM
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
johnsonandrew
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#42
Sep27-07, 09:57 PM
P: 88
Damn I was sure that was it..
johnsonandrew
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#43
Sep27-07, 09:58 PM
P: 88
I said that 30.2^2 + F^2 = 36.4^2
anglum
anglum is offline
#44
Sep27-07, 09:59 PM
P: 275
if using what u listed johnson using the pythagorean theorem the Fn = 36.0987

is that what u get as well???
anglum
anglum is offline
#45
Sep27-07, 10:00 PM
P: 275
but that is also incorrect
johnsonandrew
johnsonandrew is offline
#46
Sep27-07, 10:01 PM
P: 88
Me? No I get 20.3 or 20.4
johnsonandrew
johnsonandrew is offline
#47
Sep27-07, 10:01 PM
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
johnsonandrew
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#48
Sep27-07, 10:02 PM
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..
johnsonandrew
johnsonandrew is offline
#49
Sep27-07, 10:03 PM
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.
anglum
anglum is offline
#50
Sep27-07, 10:04 PM
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
learningphysics
learningphysics is offline
#51
Sep27-07, 10:05 PM
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?
johnsonandrew
johnsonandrew is offline
#52
Sep27-07, 10:05 PM
P: 88
dotted line is equal to parrelled force = Fg*sin x = 30.1964
johnsonandrew
johnsonandrew is offline
#53
Sep27-07, 10:07 PM
P: 88
F parallel is 30.1713, not 30.1964. AHh
johnsonandrew
johnsonandrew is offline
#54
Sep27-07, 10:07 PM
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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