```Weight & Normal Force
• Weight  The force of gravity on an object.
• Write as FG  W.
• Consider an object in free fall. Newton’s 2nd Law:
∑F = ma
• If no other forces are acting, only FG ( W) acts (in the
vertical direction).
∑Fy = may
Or:
(down, of course)
• SI Units: Newtons (just like any force!).
g = 9.8 m/s2  If m = 1 kg, W = 9.8 N
Normal Force
• Suppose an object is at rest on a table.
No motion, but does the force of gravity
stop?
OF COURSE NOT!
• But, the object does not move:
Newton’s 2nd Law is  ∑F = ma = 0
So, there must be some other force acting
besides gravity (weight) to have ∑F = 0.
• That force  The Normal Force FN (= n)
“Normal” is a math term for perpendicular ()
FN is  to the surface & equal & opposite to the weight
(true in this simple case only!) CAUTION!!
FN isn’t always = & opposite to the weight,
as we’ll see!
Example
“Free Body
 Diagram”
for the monitor.
Shows all forces
on it, in proper
directions.
Monitor at rest on table. Force of monitor on table ≡ Fmt. Force of
table on monitor ≡ Ftm. Ftm keeps monitor from falling. Ftm & Fmt
are 3rd Law action-reaction pairs. Forces on monitor are “Normal
Force” n & weight Fg. 2nd Law for monitor in vertical direction:
∑Fy = 0 = n - Fg. So, n = Fg = mg. So n = mg. They are equal &
in opposite directions, BUT THEY ARE NOT action-reaction pairs
(they act on the SAME object, not on different objects!)
Normal Force
Where does the normal force come from?
Normal Force
Where does the normal force come from?
From the other object!!!
Normal Force
Where does the normal force come from?
From the other object!!!
Is the normal force ALWAYS
equal & opposite to the weight?
Normal Force
Where does the normal force come from?
From the other object!!!
Is the normal force ALWAYS
equal & opposite to the weight?
NO!!!
An object at rest must
have no net force on it.
“Free Body
Diagrams”
If it is sitting on a table, the
force of gravity is still there;
what other force is there?
for Lincoln. Show
all forces in proper
directions.
The force exerted
perpendicular to a
surface is called the
Normal Force FN.
It is exactly as large as needed to
balance the force from the object.
(If the required force gets too big,
something breaks!)
Newton’s 2nd Law for Lincoln:
∑F = ma = 0 or FN – FG = 0 or FN = FG = mg
FN & FG AREN’T action-reaction pairs from N’s 3rd Law! They’re equal
& opposite because of N’s 2nd Law! FN & FN ARE action-reaction pairs!!
Example
m = 10 kg
l
Find: The Normal force
on the box from the
table for Figs. a., b., c.
Example
m = 10 kg
The normal
force is
NOT
l
always equal
& opposite to
the weight!!
Find: The Normal force
on the box from the
table for Figs. a., b., c.
Always use N’s 2nd
Law to CALCULATE
l
FN !
Example
m = 10 kg
The normal
force is
NOT
always equal
& opposite to
the weight!!
l
Find: The Normal force
on the box from the
table for Figs. a., b., c.
Always use N’s 2nd
Law to CALCULATE
FN !
Example
m = 10 kg
The normal
force is
NOT
always equal
& opposite to
the weight!!
Find: The Normal force
on the box from the
table for Figs. a., b., c.
Always use N’s 2nd
Law to CALCULATE
FN !
Example
l
What happens when a m = 10 kg, ∑F
= ma
person pulls upward on
F
–
mg
=
ma
P
l
the box in the previous
100 – 89 = 10a
example with a force
a = 0.2 m/s2m = 10 kg
greater than the box’s
∑F
=
ma
weight, say 100.0 N?
l The box will accelerate
upward because
FP – mg = ma
FP > mg!!
Note:
The normal force is zero here
l
because the mass isn’t in contact with
a surface!
Example
What happens when a m = 10 kg, ∑F = ma
person pulls upward on
FP – mg = ma
the box in the previous
100 – 89 = 10a
example with a force
a = 0.2 m/s2m = 10 kg
greater than the box’s
∑F
=
ma
weight, say 100.0 N?
The box will accelerate
upward because
FP – mg = ma
FP > mg!!
Note
The normal force is zero here
because the mass isn’t in contact with a surface!
Example: Apparent “weight loss”
A 65-kg (mg = 640 N) woman descends in an
elevator that accelerates at a rate a = 0.20g
downward. She stands on a scale that reads in kg.
(a) During this acceleration, what is her weight &
(b) What does the scale read when the elevator
descends at a constant speed of 2.0 m/s?
Example: Apparent “weight loss”
A 65-kg (mg = 640 N) woman descends in an
elevator that accelerates at a rate a = 0.20g
downward. She stands on a scale that reads in kg.
(a) During this acceleration, what is her weight &
(b) What does the scale read when the elevator
descends at a constant speed of 2.0 m/s?
Reasoning to get the solution using from Newton’s Laws
To use Newton’s 2nd Law for her, ONLY the forces acting on her are included.
By Newton’s 3rd Law, the normal force FN acting upward on her is equal &
opposite to the scale reading. So, the numerical value of FN is equal to the “weight”
she reads on the scale! Obviously, FN here is NOT equal & opposite
to her true weight mg!! How do we find FN? As always,
WE APPLY NEWTON’S 2ND LAW TO HER!!
Example: Apparent “weight loss”
A 65-kg (mg = 637 N) woman descends in an
elevator that accelerates at a rate a = 0.20g
downward. She stands on a scale that reads in kg.
(a) During this acceleration, what is her weight &
(b) What does the scale read when the elevator
descends at a constant speed of 2.0 m/s?
Solution
(a) Newton’s 2nd Law applied to the woman is (let down be positive!):
∑F = ma
Since a is a 1d vector pointing down, this gives: mg – FN = ma
so FN = mg - ma = m(g – 0.2g) = 0.8mg
which is numerically equal to the scale reading by Newton’s 3rd Law!!
So if she trusts the scale (& if she doesn’t know N’s Laws!), she’ll think
that she has lost 20% of her body weight!!
```
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