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Welcome back to PHY 211

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Second, there is the contact force produced by the scale acting on the person. ... The two forces acting on mass m are the normal force N acting upward and the ... – PowerPoint PPT presentation

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Title: Welcome back to PHY 211


1
Welcome back to PHY 211
  • Todays Agenda
  • Apparent Weight
  • Forces in Circular Motion

2
A person is standing on a bathroom scale while
riding an express elevator in a tall office
building. When the elevator is at rest, the
scale reads about 160 lbs. While the elevator
is moving, the reading is frequently changing,
with values ranging anywhere from about 120 lbs
to about 200 lbs. At a moment when the scale
shows the maximum reading (i.e., 200 lbs) the
elevator
1. must be going up 2. must be going down
3. could be going up or going down 4. Im not
sure.
3
Free body diagram for person
  • There are two forces acting on the person
  • First, there is the force of gravity produced by
    the earth, i.e. the weight of the person.
  • Second, there is the contact force produced by
    the scale acting on the person.

4
  • How do we know that there are no other forces on
    the person?
  • Answer The person is in contact only with the
    scale.

5
  • Motion of elevator
  • (if a )
  • Moving upward and slowing down,
  • OR
  • Moving downward and speeding up.
  • Motion of elevator
  • (if a )
  • Moving upward and speeding up,
  • OR
  • Moving downward and slowing down.

6
Applying Newtons Second Law
  • Fnet ma
  • Fnet is the sum of all the forces on the person.
    Also m is the mass of the person and a is the
    acceleration.
  • Fnet is N W, where N is the normal force
    produced by the scale and W is the weight of the
    person.
  • Combining, gives N W ma

7
Finding the Apparent Weight
  • The apparent weight Wapparent is simply the
  • normal force N. We simply solve for N to get
  • N W ma
  • Hence the apparent weight exceeds W, if a is
    positive. This occurs, for e.g. when the elevator
    is rising with increasing speed

8
Continued discussion of Apparent Weight
  • Suppose the elevator is descending but with
    increasing speed
  • Then acceleration a is negative
  • We again apply our general result
  • N W ma
  • We see that the normal force now is less than
    the actual weight W.

9
Special Case Free Fall
  • IF the elevator is in free fall, its acceleration
    is g. Then what is the apparent weight?
  • Again using N W ma, we see that N W mg
    0!!!
  • i.e. The scale will read zero.

10
Conclusions
  • Scale reads magnitude of normal force N
  • Reading on scale does not depend on velocity
  • Depends on acceleration only
  • agt0 normal force bigger
  • alt0 normal force smaller

11
A person is standing on a bathroom scale while
riding an express elevator in a tall office
building. When the elevator is at rest, the
scale reads about 160 lbs. While the elevator
is accelerating, a different reading is observed,
with values ranging anywhere from about 120 lbs
to about 200 lbs. At a moment when the scale
shows the maximum reading (i.e., 200 lbs) the
acceleration of the elevator is approximately
1. 1 m/s2 2. 2.5 m/s2 3. 5 m/s2 4. 12.5 m/s2
12
Problem to Test your understanding
  • A woman is in a descending elevator which is
    slowing down. It is losing 2.0m/s every second.
    The weight of the woman is 490N. Find her
    apparent weight on the scale on which she is
    standing.

13
Forces in circular motion
14
Review on Circular Motion
  • An object of mass m is moving with uniform speed
    v in a circle
  • The radius of the circle is r
  • What is the direction and magnitude of the
    acceleration vector?
  • Recall that the acceleration vector is directed
    radially inward
  • The magnitude is v2/r

15
Net Force on the mass
  • By Newtons second law, there must be a net
    force on the mass which is directed radially
    inward. This is due to the fact that the velocity
    vector is changing direction.
  • Applying Newtons second law, the net force
  • Fnet ma mv2/r

16
Example
  • A mass m moves on a frictionless track in a
    vertical plane with constant speed v
  • The radius of the track is r
  • At the bottom of its path, find the normal force
    N

17
Solution
  • The two forces acting on mass m are the normal
    force N acting upward and the weight force W
    acting downward
  • Hence Fnet N W
  • Equating to ma, we obtain
  • N W ma m(v2/r)
  • It follows that the normal force N is
  • N mg m(v2/r)
  • Note that the normal force exceeds the weight
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