w/A wedge.
x/Archimedes’ screw
This is an example of a simple machine, which is any mechanical
system that manipulates forces to do work. This particular ma-
chine reverses the direction of the motion, but doesn’t change the
force or the speed of motion.
A mechanical advantage example 37
The idealized pulley in figure u has negligible mass, so its kinetic
energy is zero, and the kinetic energy theorem tells us that the
total force on it is zero. We know, as in the preceding example,
that the two forces pulling it to the right are equal to each other,
so the force on the left must be twice as strong. This simple
machine doubles the applied force, and we refer to this ratio as
amechanical advantage(M.A.) of 2. There’s no such thing as
a free lunch, however; the distance traveled by the load is cut in
half, and there is no increase in the amount of work done.
Inclined plane and wedge example 38
In figure v, the force applied by the hand is equal to the one ap-
plied to the load, but there is a mechanical advantage compared
to the force that would have been required to lift the load straight
up. The distance traveled up the inclined plane is greater by a
factor of 1/sinθ, so by the work theorem, the force is smaller by
a factor of sinθ, and we have M.A.=1/sinθ. The wedge, w, is
similar.
Archimedes’ screw example 39
In one revolution, the crank travels a distance 2πb, and the water
rises by a heighth. The mechanical advantage is 2πb/h.3.2.10 Force related to interaction energy
In section 2.3, we saw that there were two equivalent ways of
looking at gravity, the gravitational field and the gravitational en-
ergy. They were related by the equation dU=mgdr, so if we knew
the field, we could find the energy by integration,U =∫
mgdr,
and if we knew the energy, we could find the field by differentiation,
g= (1/m) dU/dr.
The same approach can be applied to other interactions, for ex-
ample a mass on a spring. The main difference is that only in grav-
itational interactions does the strength of the interaction depend
on the mass of the object, so in general, it doesn’t make sense to
separate out the factor ofmas in the equation dU=mgdr. Since
F =mgis the gravitational force, we can rewrite the equation in
the more suggestive form dU=Fdr. This form no longer refers to
gravity specifically, and can be applied much more generally. The
only remaining detail is that I’ve been fairly cavalier about positive
and negative signs up until now. That wasn’t such a big problem
for gravitational interactions, since gravity is always attractive, but
it requires more careful treatment for nongravitational forces, where
we don’t necessarily know the direction of the force in advance, and172 Chapter 3 Conservation of Momentum