Recent content by intdx

I'm sorry; I don't understand what you mean. Would you mind drawing it? Would it be something like this? mg\textrm{cos(30}^\circ \textrm{)} EDIT - whoops, sorry, I edited without realizing you had posted again.

Homework Statement From http://library.thinkquest.org/10796/index.html m=1300kg t=35s r=40m Homework Equations V=\frac{2\pi r}{t} a_c=\frac{V^2}{r} F_c=ma_c F_f=\mu mg F_f=F_c \mu=\frac{a_c}{g}=\frac{F_c}{mg} The Attempt at a Solution Plugging into the formulae is pretty...

Actually, I suppose that combining the force of gravity in this case (98N) with a horizontal vector couldn't reduce the magnitude to just 49N... Is it the second option that I proposed, then? \sqrt{F_g^2\ \textrm{cos}^2\theta +F_g^2\ \textrm{cos}^2(90^\circ-\theta)}=F_g\textrm{,} so it seems...

Homework Statement From http://library.thinkquest.org/10796/index.html (#6b, 6c) The site accepts the answer 49N for #6c, but I'm not sure why. Homework Equations F_g=mg F=\sqrt{F_x^2+F_y^2} F_x=F\times \cos \theta The Attempt at a Solution The answer for #6c appears to be equal to...

Homework Statement From http://library.thinkquest.org/10796/index.html (#6) g=9.80m/s^2 Homework Equations F=ma F_f=\mu F_N F_N=mg (The site actually states the normal force to be equal to negative mass times gravitational acceleration, but with a negative value for gravitational...