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...