1. Conceptual Example 4-4: What exerts the force to move a car?

2. 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 3 rd Law, the ground pushes on the tires in the opposite direction, accelerating the car forward.

3. Helpful notation: On forces, the 1 st subscript is the object that the force is being exerted on; the 2 nd is the source. Action-Reaction Pairs act on Different Objects!

4. 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?

5. 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.

6. Sect. 4-6: Weight & Normal Force Weight  The force of gravity on an object. Write as F G  W. Consider an object in free fall. Newton’s 2 nd Law: ∑F = ma If no other forces are acting, only F G (  W) acts (in the vertical direction). ∑F y = ma y Or: (down, of course) SI Units: Newtons (just like any force!). g = 9.8 m/s 2  If m = 1 kg, W = 9.8 N

7. 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 2 nd 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 F N (= N) “Normal” is a math term for perpendicular (  ) F N is  to the surface & equal & opposite to the weight (true in this simple case only!) CAUTION!! F N isn’t always = & opposite to the weight, as we’ll see!

8. Normal Force Where does the normal force come from?

9. Normal Force Where does the normal force come from? From the other object!!!

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

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

12. 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 F N. 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 2 nd Law for Lincoln: ∑F = ma = 0 or F N – F G = 0 or F N = F G = mg F N & F G AREN’T action-reaction pairs from N’s 3 rd Law! They’re equal & opposite because of N’s 2 nd Law! F N & F N ARE action-reaction pairs!! “Free Body Diagrams” for Lincoln. Show all forces in proper directions.

13. m = 10 kg Example 4-6 Find: The Normal force on the box from the table for Figs. a., b., c. l

14. 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 2 nd Law to CALCULATE F N ! l l

15. Example 4-6 l Find: The Normal force on the box from the table for Figs. a., b., c. Always use N’s 2 nd Law to CALCULATE F N ! m = 10 kg The normal force is NOT always equal & opposite to the weight!!

16. Example 4-6 Find: The Normal force on the box from the table for Figs. a., b., c. Always use N’s 2 nd Law to CALCULATE F N ! m = 10 kg The normal force is NOT always equal & opposite to the weight!!

17. Example 4-7 m = 10 kg ∑F = ma F P – 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 F P > mg!! Note: The normal force is zero here because the mass isn’t in contact with a surface! m = 10 kg, ∑F = ma F P – mg = ma 100 – 89 = 10a a = 0.2 m/s 2 l l l l

18. Example 4-7 m = 10 kg ∑F = ma F P – 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 F P > mg!! Note The normal force is zero here because the mass isn’t in contact with a surface! m = 10 kg, ∑F = ma F P – mg = ma 100 – 89 = 10a a = 0.2 m/s 2

19. 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?

20. Reasoning to get the solution using from Newton’s Laws To use Newton’s 2 nd Law for her, ONLY the forces acting on her are included. By Newton’s 3 rd Law, the normal force F N acting upward on her is equal & opposite to the scale reading. So, the numerical value of F N is equal to the “weight” she reads on the scale! Obviously, F N here is NOT equal & opposite to her true weight mg!! How do we find F N ? As always, WE APPLY NEWTON’S 2 ND 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?

21. 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 2 nd Law applied to the woman is (let down be positive!): ∑F = ma Since a is a 1d vector pointing down, this gives: mg – F N = ma so F N = mg - ma = m(g – 0.2g) = 0.8mg which is numerically equal to the scale reading by Newton’s 3 rd 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!!

22. 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

23. 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

24. 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