1. Exam I is Monday, September 26!!

2. More on Newton’s 3rd Law

3. What exerts the force to move a car?
Conceptual Example:
What exerts the force to move a car?
Response: A common answer is that
the engine makes the car move
forward. But it is not so simple. The
engine makes the wheels go around.
But if the tires are on slick ice or
deep mud, they just spin.
 Friction is needed.

4. What exerts the force to move a car?
Conceptual Example:
What exerts the force to move a car?
 Friction is needed.
On firm ground, the tires push backward
against the ground because of friction.
By Newton’s 3rd Law, the ground
pushes on the tires in the opposite
direction, accelerating the car forward.

5. Action-Reaction Different Objects!
Helpful Notation
On forces, the 1st subscript is the object that the force is being exerted on; the 2nd is the source.
Action-Reaction
Pairs act on
Different Objects!
Figure Caption: We can walk forward because, when one foot pushes backward against the ground, the ground pushes forward on that foot (Newton’s third law). The two forces shown act on different objects.

6. Conceptual Example

7. Action-Reaction Pairs Act On Different Objects
Forces exerted BY an object DO NOT (directly) influence its motion!!

8. Action-Reaction Pairs Act On Different Objects
Forces exerted BY an object DO NOT (directly) influence its motion!!
Forces exerted ON an object (BY some other object) DO influence its motion!!

9. Action-Reaction Pairs Act On Different Objects
Forces exerted BY an object DO NOT (directly) influence its motion!!
Forces exerted ON an object (BY some other object) DO influence its motion!!
When discussing forces, use the words “BY” and “ON” carefully.

10. Weight  The force of gravity on an object.
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 is:
∑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

11. “Normal” Force “Normal” is a math term for perpendicular ()
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:
2nd Law  ∑F = ma = 0
 There must be some other force acting
besides gravity (weight) to have ∑F = 0.
That force  Normal Force FN (= N in text!)
“Normal” is a math term for perpendicular ()
FN is  to the surface & opposite to the weight
(in this simple case only!) Caution!! FN isn’t always = & opposite to the weight, as we’ll see!

12. ∑F = ma = 0 ∑F = FN – mg = 0 FN = mg
“Normal” Force
∑F = ma = 0
∑F = FN – mg = 0 or
FN = mg

13. A Very Important Conceptual Question!
“Normal” Force
∑F = ma = 0
∑F = FN – mg = 0 or
FN = mg
A Very Important Conceptual Question!
FN = mg

14. ∑F = ma = 0 ∑F = FN – mg = 0 FN = mg
“Normal” Force
∑F = ma = 0
∑F = FN – mg = 0 or
FN = mg
A Very Important Conceptual Question!
FN = mg
Which Newton’s Law of Motion tells us this?

15. ∑F = ma = 0 ∑F = FN – mg = 0 FN = mg
“Normal” Force
Newton’s 2nd Law
in vertical direction!
∑F = ma = 0
∑F = FN – mg = 0 or
FN = mg
A Very Important Conceptual Question!
FN = mg
Which Newton’s Law of Motion tells us this? Newton’s 2nd Law!
NOT Newton’s 3rd Law!!

16. Is the Normal Force FN Always Equal & Opposite to the Weight?

17. Normal Force
Is the Normal Force FN Always Equal & Opposite to the Weight?
NO!!!!!!

18. NO!!! Normal Force Where does the normal force come from?
From the other object!!!
Is the normal force ALWAYS equal & opposite to the weight?
NO!!!

19. Note! FN & FG AREN’T action-reaction pairs from N’s
An object at rest must
have no net force on it.
If it sits on a table, the force
of gravity is still there;
what other force is there?
The force exerted
perpendicular to a
surface is called the
Normal Force FN.
Free Body
Diagram
Shows all forces
in proper
directions.
FN is exactly as large as needed to balance the force from the
object. (If the required force gets too big, something breaks!)
∑F = ma = 0 or Newton’s 2nd Law for Lincoln:
FN – FG = 0 or FN = FG = mg
Note! 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 the action-reaction pairs!!

20. Always use N’s 2nd Law to CALCULATE FN!
Example
The normal
force is NOT
always equal
& opposite
to the weight!!
m = 10 kg
Find: The normal force on the box from the table in Figs. a, b, c.
Always use N’s 2nd Law
to CALCULATE FN!

21. Always use N’s 2nd Law to CALCULATE FN!
Example
The normal
force is NOT
always equal
& opposite
to the weight!!
m = 10 kg
Find: The normal force on the box from the table in Figs. a, b, c.
Always use N’s 2nd Law
to CALCULATE FN!

22. Always use N’s 2nd Law to CALCULATE FN!
Example
The normal
force is NOT
always equal
& opposite
to the weight!!
m = 10 kg
Find: The normal force on the box from the table in Figs. a, b, c.
Always use N’s 2nd Law
to CALCULATE FN!

23. The box will accelerate
Example
What happens when a person pulls
up on the box in the previous
example with a force of N?
The box will accelerate
upward because
FP > mg!!
m = 10 kg
∑F = ma. FP – mg = ma
100 – 98 = 10a
a = 0.2 m/s2

24. The box will accelerate
Example
What happens when a person pulls
up on the box in the previous
example with a force of N?
The box will accelerate
upward because
FP > mg!!
Note: The normal force is
zero in this case because the
mass isn’t in contact with a surface.
m = 10 kg
∑F = ma. FP – mg = ma
100 – 98 = 10a
a = 0.2 m/s2

25. Example: Apparent “Weight Loss”
A 65-kg woman descends in an elevator that accelerates at 0.20g (= 1.96 m/s2) downward.
She stands on a scale that reads in kg.
(a) During this acceleration, what is her weight
& what does the scale read?
(b) What does the scale read when the elevator
descends at a constant speed of 2.0 m/s?
Figure Caption: The acceleration vector is shown in gold to distinguish it from the red force vectors.
Her weight is always mg, but the normal force is 0.8 mg, which the scale reads as 52 kg.
The scale reads her true weight, or a mass of 65 kg.

26. Example: Apparent “Weight Loss”
A 65-kg woman descends in an elevator that accelerates at 0.20g (= 1.96 m/s2) downward.
She stands on a scale that reads in kg.
(a) During this acceleration, what is her weight
& what does the scale read?
(b) What does the scale read when the elevator
descends at a constant speed of 2.0 m/s?
Note: To use Newton’s 2nd Law for her, ONLY the forces
acting on her are included.
Figure Caption: The acceleration vector is shown in gold to distinguish it from the red force vectors.
Her weight is always mg, but the normal force is 0.8 mg, which the scale reads as 52 kg.
The scale reads her true weight, or a mass of 65 kg.

27. Example: Apparent “Weight Loss”
A 65-kg woman descends in an elevator that accelerates at 0.20g (= 1.96 m/s2) downward.
She stands on a scale that reads in kg.
(a) During this acceleration, what is her weight
& what does the scale read?
(b) What does the scale read when the elevator
descends at a constant speed of 2.0 m/s?
Note: 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?
Figure Caption: The acceleration vector is shown in gold to distinguish it from the red force vectors.
Her weight is always mg, but the normal force is 0.8 mg, which the scale reads as 52 kg.
The scale reads her true weight, or a mass of 65 kg.

28. Example: Apparent “Weight Loss”
A 65-kg woman descends in an elevator that accelerates at 0.20g (= 1.96 m/s2) downward.
She stands on a scale that reads in kg.
(a) During this acceleration, what is her weight
& what does the scale read?
(b) What does the scale read when the elevator
descends at a constant speed of 2.0 m/s?
Note: 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!!
Figure Caption: The acceleration vector is shown in gold to distinguish it from the red force vectors.
Her weight is always mg, but the normal force is 0.8 mg, which the scale reads as 52 kg.
The scale reads her true weight, or a mass of 65 kg.

29. Example: Apparent “Weight Loss”
Mass m = 65-kg, mg = 637 N
Acceleration a = 1.96 m/s2 down.
(a) During acceleration, what is her weight
& what does the scale read?
Due to Newton’s 3rd Law, the numerical value
of FN is equal to the “weight” reading on the scale! Obviously,
FN is NOT equal & opposite to her true weight mg!!
Figure Caption: The acceleration vector is shown in gold to distinguish it from the red force vectors.
Her weight is always mg, but the normal force is 0.8 mg, which the scale reads as 52 kg.
The scale reads her true weight, or a mass of 65 kg.

30. Example: Apparent “Weight Loss”
Mass m = 65-kg, mg = 637 N
Acceleration a = 1.96 m/s2 down.
(a) During acceleration, what is her weight
& what does the scale read?
Due to Newton’s 3rd Law, the numerical value
of FN is equal to the “weight” reading on the scale! Obviously,
FN is NOT equal & opposite to her true weight mg!!
Find FN by applying Newton’s 2nd Law to her!!
Let down be positive so up is negative: Fy = ma
 mg – FN = ma  FN = m(g – a) = (65)(9.8 – 1.96) = N
Figure Caption: The acceleration vector is shown in gold to distinguish it from the red force vectors.
Her weight is always mg, but the normal force is 0.8 mg, which the scale reads as 52 kg.
The scale reads her true weight, or a mass of 65 kg.

31. Example: Apparent “Weight Loss” (FN/g)= 52 kg = Scale Reading in kg
Mass m = 65-kg, mg = 637 N
Acceleration a = 1.96 m/s2 down.
(a) During acceleration, what is her weight
& what does the scale read?
Due to Newton’s 3rd Law, the numerical value
of FN is equal to the “weight” reading on the scale! Obviously,
FN is NOT equal & opposite to her true weight mg!!
Find FN by applying Newton’s 2nd Law to her!!
Let down be positive so up is negative: Fy = ma
 mg – FN = ma  FN = m(g – a) = (65)(9.8 – 1.96) = N
(FN/g)= 52 kg = Scale Reading in kg
= “Effective Weight”!
Figure Caption: The acceleration vector is shown in gold to distinguish it from the red force vectors.
Her weight is always mg, but the normal force is 0.8 mg, which the scale reads as 52 kg.
The scale reads her true weight, or a mass of 65 kg.

32. Example: Apparent “Weight Loss”
Mass m = 65-kg, mg = 637 N
Acceleration a = 1.96 m/s2 down.
(a) During acceleration, what is her weight
& what does the scale read?
(b) Answer part a if the elevator descends at a
constant speed of 2.0 m/s?
Part (b): Due to Newton’s 3rd Law, the
numerical value of FN is equal to the “weight” reading on
the scale! Since v = constant, acceleration a = 0.
Figure Caption: The acceleration vector is shown in gold to distinguish it from the red force vectors.
Her weight is always mg, but the normal force is 0.8 mg, which the scale reads as 52 kg.
The scale reads her true weight, or a mass of 65 kg.

33. Example: Apparent “Weight Loss”
Mass m = 65-kg, mg = 637 N
Acceleration a = 1.96 m/s2 down.
(a) During acceleration, what is her weight
& what does the scale read?
(b) Answer part a if the elevator descends at a
constant speed of 2.0 m/s?
Part (b): Due to Newton’s 3rd Law, the
numerical value of FN is equal to the “weight” reading on
the scale! Since v = constant, acceleration a = 0.
So, in this case Newton’s 2nd Law applied to her gives:
∑F = FN – mg = ma = 0
or FN = mg
Figure Caption: The acceleration vector is shown in gold to distinguish it from the red force vectors.
Her weight is always mg, but the normal force is 0.8 mg, which the scale reads as 52 kg.
The scale reads her true weight, or a mass of 65 kg.

34. Example: Apparent “Weight Loss”
Mass m = 65-kg, mg = 637 N
Acceleration a = 1.96 m/s2 down.
(a) During acceleration, what is her weight
& what does the scale read?
(b) Answer part a if the elevator descends at a
constant speed of 2.0 m/s?
Part (b): Due to Newton’s 3rd Law, the
numerical value of FN is equal to the “weight” reading on
the scale! Since v = constant, acceleration a = 0.
So, in this case Newton’s 2nd Law applied to her gives:
∑F = FN – mg = ma = 0
or FN = mg
From N’s 2nd Law, NOT from N’s 3rd Law!
Figure Caption: The acceleration vector is shown in gold to distinguish it from the red force vectors.
Her weight is always mg, but the normal force is 0.8 mg, which the scale reads as 52 kg.
The scale reads her true weight, or a mass of 65 kg.

35. Figure 3.9 These young people are experiencing apparent "weightlessness" in a special NASA plane. Notice the padding.
Figure 3-9 p62