Objectives: The students will be able to: Use the methods of vector algebra to determine the net force acting on an object. Draw an accurate free body diagram locating each of the forces acting on an object or a system of objects. Use free body diagrams and Newton's laws of motion to solve word problems.

Review http://www.youtube.com/watch?v=NYVMlmL0BPQ Newton’s First Law: Objects in motion tend to stay in motion and objects at rest tend to stay at rest unless acted upon by an unbalanced force. Newton’s Second Law: Force equals mass times acceleration (F = ma).` Newton’s Third Law: For every action there is an equal and opposite reaction.

Newton's Laws 1stlaw: Homer is large and has much mass, therefore he has much inertia. Friction and gravity oppose his motion. 2nd law: Homer’s mass x 9.8 m/s/s equals his weight, which is a force. 3rd law: Homer pushes against the ground and it pushes back.

Forces Newton’s Laws of Motion Free body diagrams Atwood device Weight Free fall Force and motion problems in 1-D Normal force Tension Free body diagrams Atwood device Static and kinetic friction Coefficients of friction Air resistance Terminal velocity

Examples of Forces A force is just a push or pull. Examples: an object’s weight tension in a rope friction attraction between an electron and proton Bodies don’t have to be in contact to exert forces on each other, e.g., gravity.

Fundamental Forces of Nature Gravity Attraction between any two bodies w/ mass Weakest but most dominant Electromagnetic Forces between any two bodies w/ charge Attractive or repulsive Weak nuclear force – responsible for radioactive decay Strong nuclear force – holds quarks together (constituents of protons and neutrons)

4-6 Weight – the Force of Gravity; and the Normal Force Weight is the force exerted on an object by gravity. Close to the surface of the Earth, where the gravitational force is nearly constant, the weight is:

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 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

“Normal” Force That force  The Normal Force FN (= N) 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  The Normal Force FN (= N) “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!

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?

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!!!

∑F = ma = 0 or 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. ∑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!!

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

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 I upward because FP > mg!! m = 10 kg, ∑F = ma FP – mg = ma 100 – 98 = 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!

Example 4-8: Apparent “weight loss” We apply Newton’s 2nd Law to her!! A 65-kg woman descends in an elevator that accelerates at 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? 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 4-17. 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.

Normal force When an object lies on a table or on the ground, the table or ground must exert an upward force on it, otherwise gravity would accelerate it down. This force is called the normal force. N In this particular case, N = mg. So, Fnet = 0; hence a = 0. m mg

Normal forces aren’t always up “Normal” means perpendicular. A normal force is always perpendicular to the contact surface. N For example, if a flower pot is setting on an incline, N is not vertical; it’s at a right angle to the incline. Also, in this case, mg > N. mg

Normal force directions Up You’re standing on level ground. You’re at the bottom of a circle while flying a loop-the-loop in a plane. Sideways A ladder leans up against a wall. You’re against the wall on the “Round Up” ride when the floor drops out. At an angle A race car takes a turn on a banked track. Down You’re in a roller coaster at the top of a loop.

Cases in which N  mg a Mass on incline Applied force acting on the mass Nonzero acceleration, as in an elevator or launching space shuttle FA N N a N mg mg mg

When does N = mg ? If the following conditions are satisfied, then N = mg: The object is on a level surface. There’s nothing pushing it down or pulling it up. The object is not accelerating vertically.

N and mg are NOT an Action-Reaction Pair! “Switch the nouns to find the reaction partner.” Earth FE m mg N Fg The dot represents the man. mg, his weight, is the force on the man due to the Earth. FE is the force on the Earth due to the man. N, the normal force, is the force on the man due to the ground. Fg is the force on the ground due to the man. The red vectors are an action-reaction pair. So are the blue vectors. Action-reaction pairs always act on two different bodies!

Box / Tension Problem 8 kg 5 kg 6 kg T1 frictionless floor T2 38 N A force is applied to a box that is connected to other boxes by ropes. The whole system is accelerating to the left. The problem is to find the tensions in the ropes. We can apply the 2nd Law to each box individually as well as to the whole system.

Box / Tension Analysis 8 kg 5 kg 6 kg T1 frictionless floor T2 38 N T1 pulls on the 8-kg box to the right just as hard as it pulls on the middle box to the left. T1 must be < 38 N, or the 8-kg box couldn’t accelerate. T2 pulls on the middle box to the right just as hard as it pulls on the 6-kg box to the left. T1 must be > T2 or the middle box couldn’t accelerate.

Free Body Diagram – system 19 kg 38 N mg N For convenience, we’ll choose left to be the positive direction. The total mass of all three boxes is 19 kg. N and mg cancel out. Fnet = m a implies a = 2.0 m/s2 Since the ropes don’t stretch, a will be 2.0 m/s2 for all three boxes.

Free Body Diagram – right box N and mg cancel out. For this particular box, Fnet = m a implies: T2 = 6a = 6(2) = 12 N. (Remember, a = 2 m/s2 for all three boxes.) 6 kg T2 N mg 8 kg 5 kg 6 kg T1 frictionless floor T2 38 N

Free Body Diagram – middle box N and mg cancel out again. Fnet = m a implies: T1 – T2 = 5a. So, T1 – 12 = 5(2), and T1 = 22 N 5 kg T2 = 12 N T1 mg N 8 kg 5 kg 6 kg T1 frictionless floor T2 38 N

Free Body Diagram – left box Let’s check our work using the left box. N and mg cancel out here too. Fnet = ma implies: 38 - 22 = ma = 8(2). 16 = 16. N 38 N T1 = 22 N 8 kg mg T2 T1 38 N 5 kg 6 kg 8 kg

Practice Problems Practice #1: A fisherman yanks a fish out of the water with an acceleration of 3.5 m/s² using very light fishing line that has a breaking strength of 28 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish? Practice #2: A 15.0-kg bucket is lowered by a rope in which there is 163 N of tension. What is the acceleration of the bucket? Is it up or down? Practice #3: An elevator (mass 6850 kg) is to be designed so that the maximum acceleration is 0.0780g. What are the maximum and minimum forces the motor should exert on the supporting cable? Practice #4: The cable supporting a 2001-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it gave the elevator without breaking?

Homework Chapter 4 Problems #s 5, 13, 15, 17

Closure Kahoot