d =
k
w
. If Jane weighs 100 pounds and sits 5 feet from the fulcrum, how far away
will Juan, who weighs 150 pounds, have to sit in order to balance Jane? Use
Jane’s data to find the constant of variation
k: 5 = k/100, so k = 500. Now
solve for Juan’s distance:
d = 500/150 = 3.33. Juan would have to sit 3 feet 4
inches from the fulcrum in order to balance Jane. Note that the constant
k was
computed from Jane’s statistics. If she were to change position or be replaced by
someone else, the value of
k would change.
Pulley systems are a series of ropes and wheels that help lift and support
heavy objects by distributing weight in multiple locations. Elevator shafts rely on
pulleys to move the cabin, and movers use pulleys to transport cumbersome or
heavy objects such as pianos into tall buildings. A 100 pound weight can feel like
a 50 pound weight when it is moved by a two-pulley system, because half the
weight is distributed at the other pulleys. As the number of pulleys in the system
increases, the amount of force needed to move the object decreases proportion-
ally. Therefore a three-pulley system needs a 33.33-pound force to move the 100
pound weight, a four-pulley system needs a 25 pound force to move the 100
pound weight, and so on. The force,
f , needed to move an object, the weight of
the object,
w, the number of pulleys needed in a system, p, are related with the
equation,
f = w/p. If the weight is constant, then the force applied varies in-
versely with the number of pulleys used.
Compound variation combines direct and indirect variation with two or more
independent variables. The gravitational force between two planets varies
directly as the product of the masses of the planets, and inversely as the square
of the distance between them:
F =
Gm
1
m
2
d
2
, where
F is the force in newtons, G
is a gravitational constant (
6.67 × 10
−11
newton-meters per square kilogram),
r
is the distance in meters between the centers of two planets, and
m
1
and
m
2
are
the mass of each planet in kilograms. The constant of variation would be differ-
ent if measurements are made in different units, such as in feet rather than meters
and pounds rather than kilograms. The formula works if one of the planets is the
earth and the other “planet” is a person high above the earth’s surface. It simpli-
fies to an inverse-variation formula for the weight of a body above the
earth:
W =
k
d
2
, where
W is the weight above the planet, d is the distance be-
tween the person and the
center of the earth, and
k is a constant. It may seem
strange that both masses have disappeared, but they are handled by the constant.
Consider a 170 pound astronaut who is 9,000 miles above the surface of the
earth. How much does he weigh at that altitude? First write the equation for his
weight at the surface of the earth. Since the radius of the earth is about 4,000
miles,
170 =
k
4,000
2
. Solving for
k yields, k = 2.72 × 10
9
. The inverse-square
formula is therefore
W =
2,720,000,000
d
2
. Using this formula with the distance
d =
13,000 miles from the center of the earth gives,
W ≈ 16.09. The astronaut would
weigh about 16 pounds.
The deflection
D of a diving board is a function of the weight W of the diver,
the length of the board
L, the elasticity E of the material making up the board,
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