Example : A 65.0 kg lady is standing on a Newton weigh scale
in an elevator. What is the apparent weight (or normal force) on
the lady if the elevator is ... a) at rest b) moving down at a constant
velocity of 4.0 m/s c) accelerating up at 4.0 m/s2 d) accelerating
down at 4.0 m/s2 e) in free fall (cable has been cut)
.a) ay = Q | F | N = m ( a y + g) = 65(0+10) =650 N
i
At rest, the apparent weight equal to the actual weight.
b) a y = 0
: r
f
1 F | N= m (ay + g) = 65(0 + 10) = 650 N
At constant velocity, the apparent weight equal to
the actual weight.
e| ;ay = + 4.0 m/s2| F | N= m (ay + g) = 65( +4+10) = 910 N
With an upward acceleration, the apparent weight is
bigger than the actual weight.
Apparent Weight t Actual Weight
••/•".• • Apparent Weight
• Actual Weight
• reading on a spring
scale when that object
is placed on it.
Force of earth's gravity
on an object
• Formula is...
Fg = mg
| F N | o r | F T | = m ( a + g)
Formula is...
Example : A 65.0 kg lady is standing on a Newton weigh scale
in an elevator. What is the apparent weight (or normal force) on
the lady if the elevator is ... a) at rest b) moving down at a constant
velocity of 4.0 m/s c) accelerating up at 4.0 m/s2 d) accelerating
down at 4.0 m/s2 e) in free fall (cable has been cut)
d) a y =-4.0m/s 2 | F | N= m (ay + g) = 65 (-4 + 10) = 390 N
With a downward acceleration, the apparent weight
is smaller than the actual weight.
e) a y = -10.0 m/s2 | F | N= m (ay + g) = 65 (-10 + 10) - 0 N
In free.fall, the apparent weight is zero. There is
apparent weightlessness.
Apparent Weight
General Apparent Weight Formula
4 Apparent Weight of an object is the reading
on a spring scale when that object is placed on
» For a stand-on Newton weigh scale
| FN | ( apparent weight) = m (ay + g)
» For a hanging Newton weigh scale
| FT | ( apparent weight) = m (ay + g)
» Note that these equations a combination of
vectors and scalars. The mass m and gravitational
field strength g are positive quantities, but a y is a
vector and could be up (+) or down (-)
It.
0
The person pulls down on the
hanging scale producing a
We Can have a hanging SCale. "reading" equal in magnitude
to the
The hanging spring
"apparent weight"
scale pulls up on the
Hanging spring
person with an equal
scale
tension force. The
For a hanging scale, the
sjze of the tension
apparent weight is equal to
force of the hanging
the
magnitude of the
scale on the person
equals the apparent
tension force |FT1 on the
weight.
object.
Apparent Weight
0
Deriving an Apparent Weight Formula
Apparent Weight of an object is the reading
pn a spring scale when that object is placed on
A person is in an elevator and is standing on a "stand-cm
Newton spring scale."
it.
The elevator could be
We can have a
The spring scale
exerts an equal
biit opposite
Normal force up
on the person.
The size of the
Normal force is
also equal to the
apparent weight.
Person exerts a force down on
<
_|
I the spring scale, producing a
Stand-On SCale."reading" inNewtons on the
scale. This reading is the
apparent weight.
Spring
scale
For a stand-on scale, the
apparent weight is equal to
the magnitude of the
normal force |FNI on the
object.
accelerating up or down.
Let's draw an FED of the person ^^^^^
Up+ A
/T\f
Down F.= mg
t
Spring
scale
F n e t Y =ma y
Vector Statement
FN + F g = may
Scalar Statement
FN -mg= may
F N = ma y + mg
| FN | is apparent weight = m (a +