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

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!)

“Free Body Diagram” for the monitor. Shows all forces on it, in proper directions.

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. 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!)

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!!

“Free Body Diagrams” for Lincoln. Showall forces in proper directions.

m = 10 kg

Example

Find: The Normal force on the box from the table for Figs. a., b., c.

l

m = 10 kg The normal force is

NOT always equal & opposite to the weight!!

Example

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

l

Example

lFind: The Normal force on the box from the table for Figs. a., b., c.

Always use N’s 2nd Law to CALCULATE

FN!

m = 10 kg The normal force is

NOT always equal & opposite to the weight!!

Example

Find: The Normal force on the box from the table for Figs. a., b., c.

Always use N’s 2nd Law to CALCULATE

FN!

m = 10 kg The normal force is

NOT always equal & opposite to the weight!!

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 100.0 N? The box will accelerate upward because FP > mg!!

Note: The normal force is zero here because the mass isn’t in contact with a surface!

m = 10 kg, ∑F = maFP – mg = ma100 – 89 = 10a

a = 0.2 m/s2

l

l

l

l

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 100.0 N? The box will accelerate upward because FP > mg!!

Note

The normal force is zero herebecause the mass isn’t in contact with a surface!

m = 10 kg, ∑F = maFP – mg = ma100 – 89 = 10a

a = 0.2 m/s2

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?

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 = 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?

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? 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!!