Recent content by leianne

yes i get it now yay and it's all thanks to you! thank you very very very much! i feel kind of embarrassed now that I've figured out what my silly mistake was :redface: i understand everything now and i better burn it to my memory yay! really thank you very very very much!

yay i get it now! thank you very very very much! yay I am just genuinely ecstatic right now and really really grateful! thank you thank you thank you!

WOAH it's 250 N! finally! that is awesome yay thank you so much! but what is the right formula then? how should i show that? again thank you very very much yay! is it simply : ma = Fp - mgsin(theta) - f ma = (Fpcostheta - Fpsintheta) - mgsintheta - f ?

Well actually the Fgx in my Fp - Fgx - f = ma is equal to mgsinθ. I just called it Fgx since it's the horizontal component of the weight. And the answer i kept getting using that formula is 204 N which means I am still missing 46 N (according from the answer at the back of my book. oh yes, my...

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?!:eek:) yay i think i got 356 N from inventing/mixing up formulas. wah...

Yes, I did. But isn't the question asking for the magnitude of the horizontal force? So shouldn't Fp also be broken down into components? Anyway is your diagram showing that F = Fp - Gx - f? I've tried it and it gave me 204 N. yay am i annoying you? I am really sorry.

And if possible can you/anyone also explain to me how my teacher got that formula? Please bear with my little understanding. :( Thank you very much.

Thank you so much! If my memory doesn't fail me, I've actually gotten 200N and 250 N at one time. It's just that i didn't understand how i got it. :smile: Anyway I've plugged it in in the formula (without Fg) but my answer still comes up to 356.25 N :confused: Why is that? I don't get it I...

Homework Statement Determine the magnitude of the horizontal force Fp required to push a block of a mass 30 kg, up along an inclined plane, with a constant acceleration of 0.8m/s2. The plane makes an angle of 25° to the horizontal and the coefficient of kinetic friction between the plane and...