Simple Nature - Light and Matter

object, we have static friction, which is less than µ s F n . As in that example, F s = mg sin θ, andF s < µ s F n , somg sin θ < µ s F n .From the same triangle, we have F n = mgcosθ, somg sin θ < µ s mg cos θ .Rearranging,θ < tan −1 µ s .(b) Both m and g canceled out, so the angle of repose would be the same on an asteroid.Solutions for Chapter 4Page 288, problem 1: The pliers are not moving, so their angular momentum remainsconstant at zero, and the total torque on them must be zero. Not only that, but each half ofthe pliers must have zero total torque on it. This tells us that the magnitude of the torque atone end must be the same as that at the other end. The distance from the axis to the nut isabout 2.5 cm, and the distance from the axis to the centers of the palm and fingers are about 8cm. The angles are close enough to 90 ◦ that we can pretend they’re 90 degrees, considering therough nature of the other assumptions and measurements. The result is (300 N)(2.5 cm)=(F )(8cm), or F =90 N.Page 294, problem 37: The foot of the rod is moving in a circle relative to the center of therod, with speed v = πb/T , and acceleration v 2 /(b/2) = (π 2 /8)g. This acceleration is initiallyupward, and is greater in magnitude than g, 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 againbefore 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 276. The key here is that the two parts of theacceleration are both independent of m and b, so the result is univeral, and it does suffice tocheck a graph in a single example. In practical terms, this tells us something about how difficultthe trick is to do. Because π 2 /8 = 1.23 isn’t much greater than unity, a hit that is just a littletoo weak (by a factor of 1.23 1/2 = 1.11) will cause a fairly obvious qualitative change in theresults. This is easily observed if you try it a few times with a pencil.Solutions for Chapter 7Page 441, problem 17:(a) Plugging in, we find√ √ √1 − w 1 − u 1 − v1 + w = 1 + u 1 + v.(b) First let’s simplify by squaring both sides.1 − w1 + w = 1 − u1 + u · 1 − v1 + v.For convenience, let’s write A for the right-hand side of this equation. We then have1 − w1 + w = A1 − w = A + Aw .936 Chapter Appendix 4: Hints and Solutions