Download Lesson #19 Apparent Weight Presentation file

Apparent Weight
Apparent Weight
• Apparent Weight of an object is the reading
on a ___________ scale when that object is
placed on it.
Apparent Weight
• Apparent Weight of an object is the reading
on a spring scale when that object is placed on
it.
Apparent Weight
• Apparent Weight of an object is the reading
on a spring scale when that object is placed on
it.
• We can have a stand-on scale.
Spring
scale
Apparent Weight
• Apparent Weight of an object is the reading
on a spring scale when that object is placed on
it.
• We can have a stand-on scale.
Spring
scale
Person exerts a force down on
the spring scale, producing a
“reading” in Newtons on the
scale. This reading is the
apparent weight.
Apparent Weight
• Apparent Weight of an object is the reading
on a spring scale when that object is placed on
it.
• We can have a stand-on scale.
What is the
reaction to this
force?
Spring
scale
Person exerts a force down on
the spring scale, producing a
“reading” in Newtons on the
scale. This reading is the
apparent weight.
Apparent Weight
• Apparent Weight of an object is the reading
on a spring scale when that object is placed on
it.
• We can have a stand-on scale.
The spring scale
exerts an equal
but opposite
Normal force up
on the person.
The size of the
normal force is
also equal to the
apparent weight.
FN
Spring
scale
Person exerts a force down on
the spring scale, producing a
“reading” in Newtons on the
scale. This reading is the
apparent weight.
Apparent Weight
• Apparent Weight of an object is the reading
on a spring scale when that object is placed on
it.
Person exerts a force down on
the spring scale, producing a
• We can have a stand-on scale.“reading” in Newtons on the
The spring scale
exerts an equal
but opposite
Normal force up
on the person.
The size of the
Normal force is
also equal to the
apparent weight.
scale. This reading is the
apparent weight.
FN
Spring
scale
For a stand-on scale, the
apparent weight is equal to
the magnitude of the
normal force ǀFNl on the
object.
Apparent Weight
• Apparent Weight of an object is the reading
on a spring scale when that object is placed on
it.
• We can have a hanging scale.
Hanging spring
scale
Apparent Weight
• Apparent Weight of an object is the reading
on a spring scale when that object is placed on
it.
• We can have a hanging scale.
Hanging spring
scale
The person pulls down on the
hanging scale producing a
“reading” equal in magnitude
to the “apparent weight”
Apparent Weight
• Apparent Weight of an object is the reading
on a spring scale when that object is placed on
it.
• We can have a hanging scale.
What is the
reaction to this
force?
Hanging spring
scale
The person pulls down on the
hanging scale producing a
“reading” equal in magnitude
to the “apparent weight”
Apparent Weight
• Apparent Weight of an object is the reading
on a spring scale when that object is placed on
it.
• We can have a hanging scale.
The hanging spring
scale pulls up on the
person with an equal
tension force. The
size of the tension
force of the hanging
scale on the person F
T
equals the apparent
weight.
Hanging spring
scale
The person pulls down on the
hanging scale producing a
“reading” equal in magnitude
to the “apparent weight”
Apparent Weight
• Apparent Weight of an object is the reading
on a spring scale when that object is placed on
it.
The person pulls down on the
hanging scale producing a
• We can have a hanging scale. “reading” equal in magnitude
The hanging spring
scale pulls up on the
person with an equal
tension force. The
size of the tension
force of the hanging
scale on the person F
T
equals the apparent
weight.
to the “apparent weight”
Hanging spring
scale
For a hanging scale, the
apparent weight is equal to
the magnitude of the
tension force ǀFTl on the
object.
Deriving an Apparent Weight Formula
Deriving an Apparent Weight Formula
• A person is in an elevator and is standing on a “stand-on
Newton spring scale.”
Spring
scale
Deriving an Apparent Weight Formula
• A person is in an elevator and is standing on a “stand-on
Newton spring scale.”
• The elevator could be
a
accelerating up or down.
Spring
scale
Deriving an Apparent Weight Formula
• A person is in an elevator and is standing on a “stand-on
Newton spring scale.”
• The elevator could be
a
accelerating up or down.
• Let’s draw an FBD of the person
Spring
scale
Deriving an Apparent Weight Formula
• A person is in an elevator and is standing on a “stand-on
Newton spring scale.”
• The elevator could be
a
accelerating up or down.
• Let’s draw an FBD of the person
Up +
Down -
a
Spring
scale
Deriving an Apparent Weight Formula
• A person is in an elevator and is standing on a “stand-on
Newton spring scale.”
• The elevator could be
a
accelerating up or down.
• Let’s draw an FBD of the person
Up +
FN
Down Fg
a
Spring
scale
Deriving an Apparent Weight Formula
• A person is in an elevator and is standing on a “stand-on
Newton spring scale.”
• The elevator could be
a
accelerating up or down.
• Let’s draw an FBD of the person
Up +
FN
Down Short-cut
formula for Fg ?
a
Spring
scale
Deriving an Apparent Weight Formula
• A person is in an elevator and is standing on a “stand-on
Newton spring scale.”
• The elevator could be
a
accelerating up or down.
• Let’s draw an FBD of the person
Up +
FN
Down Fg =mg
a
Spring
scale
Deriving an Apparent Weight Formula
• A person is in an elevator and is standing on a “stand-on
Newton spring scale.”
• The elevator could be
a
accelerating up or down.
• Let’s draw an FBD of the person
Up +
FN
a
Spring
scale
Let’s start with the equation of motion.
What is it?
Down Fg = mg
Deriving an Apparent Weight Formula
• A person is in an elevator and is standing on a “stand-on
Newton spring scale.”
• The elevator could be
a
accelerating up or down.
• Let’s draw an FBD of the person
Up +
FN
Spring
scale
a
Fnet Y = may
Down Fg = mg
Deriving an Apparent Weight Formula
• A person is in an elevator and is standing on a “stand-on
Newton spring scale.”
• The elevator could be
a
accelerating up or down.
• Let’s draw an FBD of the person
Up +
FN
a
Spring
scale
Fnet Y = may
?
Vector Statement
Down Fg = mg
Deriving an Apparent Weight Formula
• A person is in an elevator and is standing on a “stand-on
Newton spring scale.”
• The elevator could be
a
accelerating up or down.
• Let’s draw an FBD of the person
Up +
FN
Spring
scale
a
Fnet Y = may
FN + Fg = may
Down Fg = mg
Vector Statement
Deriving an Apparent Weight Formula
• A person is in an elevator and is standing on a “stand-on
Newton spring scale.”
• The elevator could be
a
accelerating up or down.
• Let’s draw an FBD of the person
Up +
FN
Down Fg = mg
Spring
scale
a
Fnet Y = may
FN + Fg = may
?
Vector Statement
Scalar Statement
Deriving an Apparent Weight Formula
• A person is in an elevator and is standing on a “stand-on
Newton spring scale.”
• The elevator could be
a
accelerating up or down.
• Let’s draw an FBD of the person
Up +
FN
Down Fg = mg
Spring
scale
a
Fnet Y = may
FN + Fg = may
FN - mg = may
Vector Statement
Scalar Statement
Deriving an Apparent Weight Formula
• A person is in an elevator and is standing on a “stand-on
Newton spring scale.”
• The elevator could be
a
accelerating up or down.
• Let’s draw an FBD of the person
Up +
FN
Down Fg = mg
Spring
scale
a
Fnet Y = may
FN + Fg = may
FN - mg = may
FN = ?
Vector Statement
Scalar Statement
Deriving an Apparent Weight Formula
• A person is in an elevator and is standing on a “stand-on
Newton spring scale.”
• The elevator could be
a
accelerating up or down.
• Let’s draw an FBD of the person
Up +
FN
Down Fg = mg
Spring
scale
a
Fnet Y = may
FN + Fg = may
FN - mg = may
FN = may + mg
Vector Statement
Scalar Statement
Deriving an Apparent Weight Formula
• A person is in an elevator and is standing on a “stand-on
Newton spring scale.”
• The elevator could be
a
accelerating up or down.
• Let’s draw an FBD of the person
Up +
FN
Down Fg = mg
a
Spring
scale
Fnet Y = may
FN + Fg = may
Vector Statement
FN - mg = may
Scalar Statement
FN = may + mg
ǀ FN ǀ is apparent weight
Deriving an Apparent Weight Formula
• A person is in an elevator and is standing on a “stand-on
Newton spring scale.”
• The elevator could be
a
accelerating up or down.
• Let’s draw an FBD of the person
Up +
FN
Down Fg = mg
a
Spring
scale
Fnet Y = may
FN + Fg = may
Vector Statement
FN - mg = may
Scalar Statement
FN = may + mg
ǀ FN ǀ is apparent weight = m (ay + g)
General Apparent Weight Formula
General Apparent Weight Formula
• For a stand-on Newton weigh scale
ǀ FN ǀ ( apparent weight ) = m (ay + g)
General Apparent Weight Formula
• For a stand-on Newton weigh scale
ǀ FN ǀ ( apparent weight ) = m (ay + g)
• For a hanging Newton weigh scale
ǀ FT ǀ ( apparent weight ) = m (ay + g)
General Apparent Weight Formula
• For a stand-on Newton weigh scale
ǀ FN ǀ ( apparent weight ) = m (ay + g)
• For a hanging Newton weigh scale
ǀ FT ǀ ( 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 ay is a
vector and could be up (+) or down (-)
Apparent Weight ǂ Actual Weight
• Apparent Weight
• Actual Weight
Apparent Weight ǂ Actual Weight
• Apparent Weight
• reading on a spring
scale when that object
is placed on it.
• Actual Weight
Apparent Weight ǂ Actual Weight
• Apparent Weight
• reading on a spring
scale when that object
is placed on it.
• Actual Weight
• Force of earth’s gravity
on an object
Apparent Weight ǂ Actual Weight
• Apparent Weight
• reading on a spring
scale when that object
is placed on it.
• Formula is…
ǀ FN ǀ or ǀ FT ǀ = m (ay + g)
• Actual Weight
• Force of earth’s gravity
on an object
Apparent Weight ǂ Actual Weight
• Apparent Weight
• reading on a spring
scale when that object
is placed on it.
• Formula is…
ǀ FN ǀ or ǀ FT ǀ = m (ay + g)
• Actual Weight
• Force of earth’s gravity
on an object
• Formula is…
Fg = m g
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)
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 = ?
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 = 0
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 = 0 ǀ FN ǀ = m (ay + g) = ?
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 = 0 ǀ FN ǀ = m (ay + g) = 65(0 +10) = ?
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 = 0 ǀ FN ǀ = m (ay + g) = 65(0 +10) =650 N
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 = 0 ǀ FN ǀ = m (ay + g) = 65(0 +10) =650 N
At rest, the apparent weight equal to the actual weight.
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 = 0 ǀ FN ǀ = m (ay + g) = 65(0 +10) =650 N
At rest, the apparent weight equal to the actual weight.
b) ay = ?
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 = 0 ǀ F ǀ N= m (ay + g) = 65(0 +10) =650 N
At rest, the apparent weight equal to the actual weight.
b) ay = 0
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 = 0 ǀ F ǀ N= m (ay + g) = 65(0 +10) =650 N
At rest, the apparent weight equal to the actual weight.
b) ay = 0 ǀ F ǀ N= m (ay + g) = ?
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 = 0 ǀ F ǀ N= m (ay + g) = 65(0 +10) =650 N
At rest, the apparent weight equal to the actual weight.
b) ay = 0 ǀ F ǀ N= m (ay + g) = 65(0 + 10) = ?
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 = 0 ǀ F ǀ N= m (ay + g) = 65(0 +10) =650 N
At rest, the apparent weight equal to the actual weight.
b) ay = 0 ǀ F ǀ N= m (ay + g) = 65(0 + 10) = 650 N
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 = 0 ǀ F ǀ N= m (ay + g) = 65(0 +10) =650 N
At rest, the apparent weight equal to the actual weight.
b) ay = 0 ǀ F ǀ N= m (ay + g) = 65(0 + 10) = 650 N
At constant velocity, the apparent weight equal to
the actual weight.
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 = 0 ǀ F ǀ N= m (ay + g) = 65(0 +10) =650 N
At rest, the apparent weight equal to the actual weight.
b) ay = 0 ǀ F ǀ N= m (ay + g) = 65(0 + 10) = 650 N
At constant velocity, the apparent weight equal to
the actual weight.
c) ay = ?
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 = 0 ǀ F ǀ N= m (ay + g) = 65(0 +10) =650 N
At rest, the apparent weight equal to the actual weight.
b) ay = 0 ǀ F ǀ N= m (ay + g) = 65(0 + 10) = 650 N
At constant velocity, the apparent weight equal to
the actual weight.
c) ay = + 4.0 m/s2
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 = 0 ǀ F ǀ N= m (ay + g) = 65(0 +10) =650 N
At rest, the apparent weight equal to the actual weight.
b) ay = 0 ǀ F ǀ N= m (ay + g) = 65(0 + 10) = 650 N
At constant velocity, the apparent weight equal to
the actual weight.
c) ay = + 4.0 m/s2 ǀ F ǀ N= m (ay + g) = ?
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 = 0 ǀ F ǀ N= m (ay + g) = 65(0 +10) =650 N
At rest, the apparent weight equal to the actual weight.
b) ay = 0 ǀ F ǀ N= m (ay + g) = 65(0 + 10) = 650 N
At constant velocity, the apparent weight equal to
the actual weight.
c) ay = + 4.0 m/s2 ǀ F ǀ N= m (ay + g) = 65( +4+10) = ?
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 = 0 ǀ F ǀ N= m (ay + g) = 65(0 +10) =650 N
At rest, the apparent weight equal to the actual weight.
b) ay = 0 ǀ F ǀ N= m (ay + g) = 65(0 + 10) = 650 N
At constant velocity, the apparent weight equal to
the actual weight.
c) ay = + 4.0 m/s2 ǀ F ǀ N= m (ay + g) = 65( +4+10) = 910 N
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 = 0 ǀ F ǀ N= m (ay + g) = 65(0 +10) =650 N
At rest, the apparent weight equal to the actual weight.
b) ay = 0 ǀ F ǀ N= m (ay + g) = 65(0 + 10) = 650 N
At constant velocity, the apparent weight equal to
the actual weight.
c) ay = + 4.0 m/s2 ǀ F ǀ N= m (ay + g) = 65( +4+10) = 910 N
With an upward acceleration, the apparent weight is
bigger than the actual weight.
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) ay = ?
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) ay = - 4.0 m/s2
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) ay = - 4.0 m/s2 ǀ F ǀ N= m (ay + g) = ?
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) ay = - 4.0 m/s2 ǀ F ǀ N= m (ay + g) = 65 (-4 + 10) = ?
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) ay = - 4.0 m/s2 ǀ F ǀ N= m (ay + g) = 65 (-4 + 10) = 390 N
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) ay = - 4.0 m/s2 ǀ F ǀ N= m (ay + g) = 65 (-4 + 10) = 390 N
With a downward acceleration, the apparent weight
is smaller than the actual weight.
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) ay = - 4.0 m/s2 ǀ F ǀ N= m (ay + g) = 65 (-4 + 10) = 390 N
With a downward acceleration, the apparent weight
is smaller than the actual weight.
e) ay = ?
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) ay = - 4.0 m/s2 ǀ F ǀ N= m (ay + g) = 65 (-4 + 10) = 390 N
With a downward acceleration, the apparent weight
is smaller than the actual weight.
e) ay = - 10.0 m/s2
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) ay = - 4.0 m/s2 ǀ F ǀ N= m (ay + g) = 65 (-4 + 10) = 390 N
With a downward acceleration, the apparent weight
is smaller than the actual weight.
e) ay = - 10.0 m/s2 ǀ F ǀ N= m (ay + g) = ?
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) ay = - 4.0 m/s2 ǀ F ǀ N= m (ay + g) = 65 (-4 + 10) = 390 N
With a downward acceleration, the apparent weight
is smaller than the actual weight.
e) ay = - 10.0 m/s2 ǀ F ǀ N= m (ay + g) = 65 (-10 + 10) = ?
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) ay = - 4.0 m/s2 ǀ F ǀ N= m (ay + g) = 65 (-4 + 10) = 390 N
With a downward acceleration, the apparent weight
is smaller than the actual weight.
e) ay = - 10.0 m/s2 ǀ F ǀ N= m (ay + g) = 65 (-10 + 10) = 0 N
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) ay = - 4.0 m/s2 ǀ F ǀ N= m (ay + g) = 65 (-4 + 10) = 390 N
With a downward acceleration, the apparent weight
is smaller than the actual weight.
e) ay = - 10.0 m/s2 ǀ F ǀ N= m (ay + g) = 65 (-10 + 10) = 0 N
In free fall, the apparent weight is zero. There is
apparent weightlessness.