physics problem :D
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Off Topic physics problem :D180 is upside down....
with that it could be even 90°
a. The net force in the horizontal direction of the system is zero
b. The net force in the vertical direction of the system is also zero
c. Since we don't want the ladder to rotated either, the torsional force along the axis of rotation normal to the xy plane must be zero as well.
Orientation:
Assume these definitions:
fg - friction on the ground
ng - normal force on the ground
w - weight
fw - friction on the wall
nw - normal on the wall
l - length of the ladder
Taking conditions a. and b. into consideration, we have:
a) x-direction: fg - nw = 0
b) y-direction: ng + fw - w = 0
For c, we need to pick some random pivot point and calculate the torque. For simplicity's sake, I chose the contact point between the ladder and the ground:
Components:
a. From ng: 0
b. From fg: 0
c. From w (assuming cm is at the exact middle of the ladder): -l/2 sin(theta) * w
d. From nw: l cos(theta) * nw
e. From fw: l sin(theta) * fw
(We note that a <- and a ^ direction force causes the same rotation, so we arbitrarily chose the torque cased by the weight in the downward direction to be negative)
So we have our final bound:
l(-w/s sin(t) + nw cos(t) + fw sin(t)) = 0
casting out the l, we have
c) -(w/2 - fw) sin(t) + nw cos(t) = 0
Now, friction = normal force * u, so doing a bunch of algebra work, we get
tan(theta) = nw/(nw*(1/2ug - uw/2))
so theta = arctan(1 / (1/(2ug) - uw/2))
I may have made some algebraic mistakes though.
wait lemme check
edit:
ya that's correct gj
now figure out part b
PS: the weight is assumed to be known and we need to find the normal forces?
2× editiert, zuletzt 27.05.11 04:12:26
hint: 4 unknown forces, 3 equations from standard methods (2 components of force + torque)