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!
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...
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.
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...