Rearranging,
θ <tan−^1 μs.
(b) Bothmandgcanceled out, so the angle of repose would be the same on an asteroid.
Page 242, problem 88:
(a) Based on units, we must haveg=kGλ/y, wherekis a unitless universal constant.
(b) For the actual calculation, we haveg=∫
dgy=G
∫
dm
r^2
cosθ,whereθis the angle between the perpendicular and thervector. Then dm=λdx, cosθ=y/r,
andr=
√
x^2 +y^2 , sog=G∫∞
−∞λdx
x^2 +y^2·
b
√
x^2 +y^2=Gλy∫∞
−∞(x^2 +y^2 )−^3 /^2 dx.Even though this has limits of integration, this is an indefinite integral because it contains the
variabley. It’s nicer to clean this up by doing a change of variable to the unitless quantity
u=x/y, giving
g=
Gλ
y∫∞
−∞(u^2 + 1)−^3 /^2 du.The definite integral is the sort of thing that sane people these days will do using computer
software. It equals 2. The result for the field is
g=
2 Gλ
y.
Solutions for chapter 4
Page 294, problem 1:
The pliers are not moving, so their angular momentum remains constant at zero, and the total
torque on them must be zero. Not only that, but each half of the pliers must have zero total
torque on it. This tells us that the magnitude of the torque at one end must be the same as
that at the other end. The distance from the axis to the nut is about 2.5 cm, and the distance
from the axis to the centers of the palm and fingers are about 8 cm. The angles are close
enough to 90◦that we can pretend they’re 90 degrees, considering the rough nature of the other
assumptions and measurements. The result is (300 N)(2.5 cm) = (F)(8 cm), orF= 90 N.
Page 301, problem 37:
The foot of the rod is moving in a circle relative to the center of the rod, with speedv=πb/T,
and accelerationv^2 /(b/2) = (π^2 /8)g. This acceleration is initially upward, and is greater in
magnitude thang, so the foot of the rod will lift off without dragging. We could also worry
about whether the foot of the rod would make contact with the floor again before the rod
finishes up flat on its back. This is a question that can be settled by graphing, or simply by
inspection of figure i on page 282. The key here is that the two parts of the acceleration are