Quote by tinytim
(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 F _{p} 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?