Download Section 2.10: Apparent Weight

Confirming Pages
1
2-1
Section 2.10: Apparent Weight
Jump off a diving board into a pool
and you feel “weightless” until you
hit the water. In reality, a gravitational
force is acting on you during your fall.
What is missing is the upward reaction force provided by whatever supports your weight at other times—the
seat of a chair, the floor of a room, the
diving board, the water.
It is useful to distinguish between
the actual weight of an object, which
is the gravitational force that acts on
it, and its apparent weight, which is
the force the object exerts on whatever it rests on. You can think of your
apparent weight as the reading of a
bathroom scale you are standing on.
If you are falling freely, the scale is
a=0
dropping just as fast as you are, and
your apparent weight is then zero.
Suppose you are standing on a
scale in an elevator, as in Figure 1.
When the elevator’s upward acceleration is a, the upward force FY on
you is the sum of your actual weight
mg and the force that accelerates you
upward, which is ma. Hence
Fy 5 mg 1 ma
The reaction force to Fy has the same
magnitude and is the force you exert
on the scale, which is your apparent
weight wapp. Hence
wapp 5 mg 1
ma
Apparent
Apparent
Actual
weight
weight
weight
a
700 N
At rest or constant
velocity
a
800 N
Upward
acceleration
If the acceleration a is positive, corresponding to an upward acceleration, then wapp > mg, and the scale
reading will be greater than your
actual weight. If a is negative, corresponding to a downward acceleration, wapp < mg, and the scale reading
will be less than your actual weight.
If the cable that supports the elevator breaks and the elevator falls freely,
a 5 2g and wapp 5 mg 2 mg 5 0.
When the elevator is at rest or moving
at constant speed up or down, a 5 0
and wapp 5 mg.
g
600 N
Downward
acceleration
0
Free fall
FIGURE 1
The actual weight of this person is 700 N. When the elevator is accelerated
upward, her apparent weight is greater, and when the elevator is accelerated downward, her
apparent weight is smaller. In free fall, her apparent weight is zero.
kra1392X_ch02_online.indd A-1
17/12/12 6:29 PM
Confirming Pages
2
2-2
Section 2.10 Example
You are standing on a scale in an elevator. When the elevator is at rest on the fourteenth
floor, the scale reads 700 N. When the elevator starts to move, the scale reads 600 N.
Describe the motion of the elevator.
Solution
Here your actual weight is 700 N and your apparent weight is 600 N. Because wapp and mg
are different, the elevator must be accelerating, and because wapp < mg, the acceleration
must be negative, that is, downward.
Your mass is
mg ________
700 N
m 5 ___
g 5 9.80 m/s2 5 71.4 kg
Solving the apparent-weight equation for the elevator’s acceleration gives
wapp
600 N
2
2
2
______
a 5 ____
m 2 g 5 71.4 kg 2 9.80 m/s 5 (8.40 2 9.80) m/s 5 2 1.40 m/s
Section 2.12: Example
A road has a hump 12 m in radius. What is the minimum speed at which a car will leave
the road at the top of the hump?
Solution
The car will leave the hump when the required centripetal force mv2/r is more than the
car’s weight of mg, since it is this weight that is providing the centripetal force. Hence
2
mv ,
mg 5 ____
r
2
v,
g 5 __
r
__
v 5 √ rg
We note that the mass of the car does not matter here. (We will find the same formula later
as the orbital speed a satellite of the earth must have.) Substituting for r and g gives
______________
v 5 √ (12 m)(9.8 m/s2) 5 10.8 m/s
This is about 24 mi/h. Driving faster over the hump is not recommended.
kra1392X_ch02_online.indd A-2
17/12/12 6:29 PM