Conceptual Example 4-4:

What exerts the force to move a car?

Conceptual Example 4-4:

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. 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.

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 onDifferent Objects!

Conceptual Example 4-5

Michelangelo’s assistant has been assigned the task of moving a block of marble using a sled. He says to his boss, “When I exert a forward force on the sled, the sled exerts an equal and opposite force backward. So how can I ever start it moving? No matter how hard I pull, the backward reaction force always equals my forward force, so the net force must be zero. I’ll never be able to move this load.” Is he correct?

Action-Reaction PairsAct 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.

Sect. 4-6: 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!

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 4-6

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 4-6

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 4-6

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 4-6

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 4-7

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

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Example 4-7

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 4-8: 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 4-8: 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 4-8: 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!!

Conceptual Example 1

• Expt 1: Cart on air track at rest. Constant force acts for short time & gives cart a certain speed.

• Expt 2: Same conditions, but force half as large as in Expt 1. Question: To reach same final speed as Expt 1, this force must act for a time which is:1. 4 times as long as in Expt 1, 2. 4 times as long as in Expt 1, 3. Equal to the time in Expt 1, 4. Half as long as time in Expt 1, 5. A quarter as long as in Expt 1

Conceptual Example 2

• Expt 1: Cart on air track at rest. Constant force acts for short time & gives cart a certain speed.

• Expt 3: Another cart, twice the mass of that in Expt. 1. Same force for same time as in Expt 1.

• Question: Final speed of heavier cart is:1. One fourth that in Expt 1. 2. 4 times that in Expt 1, 3. Half that in Expt 1, 4. Double that in Expt 1, 5. Same as in Expt 1

Conceptual Example 3

• Expt 1: Cart on air track at rest. Constant force acts for short time & gives cart a certain speed.

• Expt 4: Same cart, but initially cart is already moving in direction of force. Exert same force for same time as in Expt. 1.

• Question: Increase in speed of cart is: 1. Equal to 2 times initial speed. 2. Equal to square of initial speed. 3. Equal to 4 times initial speed. 4. Same as in Expt 1, 5. Can’t be determined from info provided