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leianne
leianne is offline
#9
Mar5-12, 05:58 AM
P: 9
Quote Quote by tiny-tim View Post
(tonit, your diagram is clearly wrong, that is not being helpful )

hi leianne!


can you show us exactly how you got from that formula to your answer?



your teacher did F = ma along the slope

really, all the forces should be on the left, and only ma on the right

(in my opinion, it is really confusing to refer to "acceleration force" )

so, on the left, it is obvious that Fp is positive, but the other two are negative

Fpcosθ = μ(mgcosθ + Fpsinθ) + Fa
Fpcos(25) = 0.21[(30x9.81xcos25) + Fpsin(25)] + 30x0.8
0.9Fp = 0.21(266.7 + 0.4Fp) + 24
0.9Fp = 56 + 0.084Fp + 24
0.9Fp - 0.084Fp = 56 + 24
0.816Fp/0.816 = 80/0.816
Fp = 98 N (what?!)

yay i think i got 356 N from inventing/mixing up formulas. wah what should i do?

anyway with what you last said, will the formula then be :

Fp - Fgx - f = ma?