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Magnitude of Force within incline?

by anglum
Tags: force, incline, magnitude
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anglum
#55
Sep27-07, 10:08 PM
P: 275
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
#56
Sep27-07, 10:09 PM
HW Helper
P: 4,125
Quote Quote by johnsonandrew View Post
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
#57
Sep27-07, 10:10 PM
P: 275
Normal force then equals 24.4307N???
learningphysics
#58
Sep27-07, 10:11 PM
HW Helper
P: 4,125
Quote Quote by anglum View Post
Normal force then equals 24.4307N???
no. how did you get that?
johnsonandrew
#59
Sep27-07, 10:13 PM
P: 88
24.4 N?
anglum
#60
Sep27-07, 10:13 PM
P: 275
Fn -53.9cos(34) -20.3 = 0
Fn = 53.9cos(34) + 20.3
Fn = 44.685 + 20.3
Fn = 64.9851N
learningphysics
#61
Sep27-07, 10:14 PM
HW Helper
P: 4,125
Quote Quote by anglum View Post
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
#62
Sep27-07, 10:15 PM
P: 88
Fn- 54cos(34) - 20.3 = 0
Fn - 44.7 - 20.3= 0 <<----
Fn= 24.4

Edit: Nevermind I know what I did...
anglum
#63
Sep27-07, 10:15 PM
P: 275
ok that is the correct answer

so our problem was we werent 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
#64
Sep27-07, 10:16 PM
P: 88
I'm so confused... I'm gonna go read over all that again haha..
learningphysics
#65
Sep27-07, 10:18 PM
HW Helper
P: 4,125
Quote Quote by anglum View Post
ok that is the correct answer

so our problem was we werent 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 [tex]\Sigma{Fy} = 0[/tex], where y is perpendicular to the plane... plug in your forces into the left side... Fn - mgcos(34) - Fsin(34) = 0...
johnsonandrew
#66
Sep27-07, 10:19 PM
P: 88
Ohhh
johnsonandrew
#67
Sep27-07, 10:20 PM
P: 88
So Fn - Fg parallel - Fg perpendicular = 0 ?
anglum
#68
Sep27-07, 10:20 PM
P: 275
thanks to all who have helped me .... tonite i finished earlier than usual... i got the other 9 problems sumwhat easily... thanks again
johnsonandrew
#69
Sep27-07, 10:21 PM
P: 88
Thanks for posting it I learned something too. Thanks learningphysics..


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