Help with AP Physics problems about motion and equilibrium?

Question

I am having difficulty with these problems, please show how to solve these. Thank you!

1) An 80kg man is one-fourth of the way up a 10m ladder that is resting against a smooth, frictionless wall. If the ladder has a mass of 20kg and it makes an angle of 60 degrees with the ground, find the force of friction of the ground on the foot of the ladder.

2) Refer to this picture...

http://img124.imageshack.us/my.php?image=physlp7.jpg

Two identical signs, each of mass M, are held aloft in adjacent archways. Following a large gust of wind, several ropes holding the signs snap, leaving only those shown in the picture. If each sign is in translational equilibrium, find the tensions of the remaining ropes in terms of its mass and gravity, g.
T1 has a 30 degree angle, T2 is 90, T3 is 45, T4 is 60.

Answer 1

for the ladder, summing torque at the foot to find the reaction force at the top, Fx

9.81*cos(60)*(2.5*80+5*20)-
Fx*sin(60)*10=0

solve for Fx
Fx=170

since this is purely horizontal and the only horizontal that is balanced by friction, this is the magnitude of the frictional force.

For the signs:

The one with T1 and T2.
The vertical component of T1 must support the weight of the sign, so
T1*sin(30)=M*g
Solve for T1
T1=M*g/sin(30)

T2 is equal to the horizontal component of T1
T2=Cos(30)*M*g/sin(30)
or
T2=M*g/Tan(30)

for the other sign, this will require looking at x and y simultaneously

Y:
T3*sin(45)+M*g=T4*sin(60)
X:
T3*cos(45)=T4*cos(60)
divide
(T3*sin(45)+M*g)/cos(45)=tan(60)
solve for T3 and then T4

T3=(cos(45)*tan(60)-M*g)/sin(45)

T4=(cos(45)*tan(60)-M*g)/
(Tan(45)*cos(60))


j

Answer 2

a million and a pair of are conservation of angular momentum. L = I * w the place L = angular momentum, I = 2d of inertia, and w = omega = angular velocity. a million) Is the rod massless? I of one weight in beginning up = m * r^2 (a formula) = 35 * .5^2 = 35/4 kg * m^2 so the internet I in beginning up = 2 * (I of one weight) = 35/2 kg * m^2 I of one weight in end = m * r^2 (a formula) = 35 * a million^2 = 35 kg * m^2 so the internet I in end = 2 * (I of one weight) = 70 kg * m^2 now, you already know w interior the beginning up = 12 rad/s so L in beginning up = L in end I * w = I * w 35/2 * 12 = 70 * w 70 * 3 = 70 * w w = 3 rad/s 2) same issues: L in beginning up = L in end Io * wo = If * wf the place o = knot = beginning up, and f = end If/Io = wo/wf = 2.5/6 = 5/12 Now we could examine Kinetic potential, ok. formula is: ok = .5 * I * w^2 and we want Kf / Ko (.5 * If * wf^2) / (.5 * Io * wo^2) .5 cancel, and replace (If * 6^2) / (Io * (5/2)^2) now we sparkling up: If/Io * (36 / (25/4) ) 5 /12 * (36 * 4 / 25 ) 3 * 4 / 5 12/5 = 2.4 3)For this we are in a position to take a glance at a pair of issues: F = ma and torque = I * alpha. Or potential. i'm going to tutor potential, yet you are able to the two, consistent with you given, i've got faith potential is far less annoying here. gravitation ability potential = kinetic rotation + translational potential Ug = Kr + Kt Ug = m*g*h Kr = .5 * I * w^2 Kt = .5 * m * v^2 and, all of us understand that v = w * r. so we are able to replace w = v/r m*g*h = .5(.5 * m * r^2 )*(v/r)^2 + .5 * m * v^2 hundreds cancel, and we in basic terms sparkling up. g*h = .25 * v^2 + .5 * v^2 g * h = .seventy 5 * v^2 = 3/4 * v^2 4 * g * h / 3 = v^2 v^2 = 4 * 9.8 * a million.2 / 3 v^2 = 15.sixty 8 v = 3.ninety six that's close sufficient to 4. so v = 4 m/s