2. 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

3. Normal Force FN isn’t always = & opposite to the weight, as we’ll see!
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!

4. Example “Free Body  Diagram” for the monitor.
Shows all forces
on it, in proper
directions.
Figure 5.6: (a) When a computer monitor is at rest on a table, the forces acting on the monitor are the normal force n and the gravitational force Fg . The reaction to n is the force Fmt exerted by the monitor on the table. The reaction to Fg is the force FmE exerted by the monitor on the Earth. (b) The free-body diagram for the monitor.
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!)

5. Where does the normal force come from?

6. Where does the normal force come from?
From the other object!!!

7. Is the normal force ALWAYS equal & opposite to the weight?
Where does the normal force come from?
From the other object!!!
Is the normal force ALWAYS equal & opposite to the weight?

8. NO!!! 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?
NO!!!

9. Newton’s 2nd Law for Lincoln:
An object at rest must have no net force on it If it is sitting 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 It is exactly as large as needed to balance the force from the object. (If the required force gets too big, something breaks!)
“Free Body
Diagrams”
for Lincoln. Show
all forces in proper
directions.
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!!

10. Example
m = 10 kg l
Find: The Normal force on the box from the table for Figs. a., b., c.

11. 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! l

12. 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!

13. 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!

14. Example m = 10 kg ∑F = ma FP – mg = ma
What happens when a person pulls upward on the box in the previous example with a force greater than the box’s weight, say N? The box will accelerate upward because
FP > mg!!
m = 10 kg, ∑F = ma
FP – mg = ma
100 – 89 = 10a
a = 0.2 m/s2 l
m = 10 kg
∑F = ma
FP – mg = ma l l
Note: The normal force is zero here
because the mass isn’t in contact with
a surface!

15. Example m = 10 kg ∑F = ma FP – mg = ma
What happens when a person pulls upward on the box in the previous example with a force greater than the box’s weight, say N? The box will accelerate upward because
FP > mg!!
m = 10 kg, ∑F = ma
FP – mg = ma
100 – 89 = 10a
a = 0.2 m/s2
m = 10 kg
∑F = ma
FP – mg = ma
Note
The normal force is zero here
because the mass isn’t in contact with a surface!

16. 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 & 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.

17. WE APPLY NEWTON’S 2ND LAW TO HER!!
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 & 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.
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!!

18. 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 & 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.
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!!