Download Example 4-7 How to “Lose Weight” the Easy Way

Example 4-7 How to “Lose Weight” the Easy Way
A 70.0-kg student stands on a bathroom scale in an elevator. What is the reading on the scale (a) when the elevator is
moving upward at a constant 0.500 m>s? (b) When the elevator is accelerating downward at 1.00 m>s 2?
Set Up
The reading on a bathroom scale tells you
how hard you are pushing on it, which may
not be the same as your weight. (Try pushing
on a bathroom scale with your hand. The
reading measures how strong you are, not
how much you weigh.) Newton’s third law
tells us that the downward normal force of the
student on the scale is equal in magnitude to
the upward normal force exerted by the scale
sscale on student. So, what we
on the student, n
sscale on student
want to know is the magnitude of n
in each situation, which we’ll find by applying
Newton’s second law to the student. The
only other force acting on the student is the
s student. Both forces and
gravitational force w
the acceleration in part (b) are in the vertical
direction, so we only need the y component.
Solve
(a) If the elevator is moving at a constant
velocity, the student’s acceleration is zero. So,
the sum of the external forces on the student
must be zero. We then solve for the magnitude
of the normal force that the scale exerts on the
student.
(b) If the elevator, and so also the student,
are accelerating downward (in the negative y
direction) at 1.00 m>s 2, the net force on the
student must be downward. The net force can
only be downward if the upward normal force
exerted by the scale is smaller than the downward gravitational force. Again we solve for
the magnitude nscale on student using Newton’s
second law.
y component of Newton’s second law
applied to the student:
nscale on student
a Fext on student, y = mstudent astudent, y (4-4b)
y
Net force on student = sum of normal
force and gravitational force:
x
s
sscale on student + w
s student
a Fext on student = n
wstudent
Student’s y velocity is constant, so
astudent, y = 0
constant velocity: astudent, y = 0
Newton’s second law in component
form:
y: a Fext on student, y
= nscale on student + (2wstudent)
= mstudent astudent, y = 0 so
nscale on student
= wstudent = mstudent g
= 170.0 kg2 19.80 m>s 2 2
= 686 N
nscale on student
net external force = 0
y
x
The student’s acceleration has a
downward (negative) y component:
astudent, y = -1.00 m>s 2
Newton’s second law in component
form:
wstudent
downward acceleration
nscale on student
y
net external force
is downward
y: a Fext on student, y
= nscale on student + (2wstudent)
astudent, y < 0
= mstudent astudent, y so
nscale on student
= wstudent + mstudent astudent, y
= 686 N + 170.0 kg2 1 -1.00 m>s 2 2 = 616 N
x
wstudent
Reflect
According to the scale, the student “weighs”
70 N less when the elevator accelerates
downward. You’ve probably experienced this
phenomenon in an elevator when it starts moving downward from rest, or when it comes to
a halt when moving upward. A similar phenomenon occurs in an elevator that accelerates
upward. Then astudent, y is positive, nscale on student
is greater than the student’s weight wstudent, and
the student feels heavier than normal.
What if the elevator is in free fall? Then
astudent, y = -g = -9.80 m>s 2 (negative because
the acceleration is downward). Substituting this
value into our result gives nscale on student = 0. The
scale reads zero, and the student will feel weightless! This situation is called apparent weightlessness. (The student isn’t truly weightless because
the gravitational force still acts on her.) We’ll see
in Chapter 10 that astronauts in Earth orbit are
also in free fall, so they feel weightless just like
the student in the free-falling elevator.
If the elevator is in free fall,
astudent, y = 2g
nscale on student
= wstudent + mstudent astudent, y
= mstudent g + mstudent(2g)
=0
free fall
normal force = 0
y
x
astudent, y = –g
wstudent