Chapter 4 Sec 6-8

1. Chapter 4 Sec 6-8 Weight, Vector Components, and Friction

2. The Force of Gravity, Weight, and the Normal Force • The force causing all objects to accelerate toward earth’s is called the force of gravity, g. • Applying Newton’s Second Law, F=ma, we obtain the force of gravity on an object accelerating at 9.80m/s/s is called Weight, Fg=mg and is measured in Newtons, N. • When two objects are in contact with each other a contact force acts perpendicular to the surface of contact and is called a Normal force, FN.

3. Do forces cancel? • Newton’s third law of motion describes ‘action-reaction’ pairs of forces. It is important to note these forces are on DIFFERENT objects, so they do NOT cancel. (Example: carry a ball: ball on back force,F1, does not cancel back on ball force,F2, because F1 is on the back and F2 is on the ball. • Normal force, FN on an object is not cancelled by Earth’s Fg on the same object.

4. Example: Consider a box on a table  • A friend gives you a gift box with a mass of 10.0 kg. The box rests on a horizontal table. Determine the weight of the box and the normal force on it. If your friend pushes down on the box with a force of 40.0 N, determine the new normal force on the box. Be sure to draw free body diagrams to show the forces acting.

5. Solution • Given: m = 10.0 kg, g = 9.80m/s/s, FN = ? • Fg = mg = 10.0 kg * 9.80m/s/s = 98 N acting downward. The only other force on the box is the normal force, FN and acts upward. • Since the box is at rest and the upward direction is positive, the sum of the y forces =0 so FN = mg. • The normal force on the box, exerted by the table, is 98.0N upward, and has a magnitude equal to the box’s weight.

6. Solution part B • If your friend pushes down with 40.0N of force, the normal force will change to keep the net force = 0 (assuming the box doesn’t move) • Now there are 3 forces on the box. • The weight is still 98.0 N (mg). The net force, ΣFy=0, so FN-mg-40.0N = 0 and FN = 40.0N + mg = 98.0N + 40.0 N = 138.0 N so the table pushes back with more force than before.

7. Example 2 Accelerating the box • What happens when a person pulls upward on the box in the previous problem with a force equal to or greater than the box’s weight, say Fp = 100.0 N? • The net force is now • With a net force larger than zero, the box will accelerate upward.

8. Combining Force Vectors • The net force on an object is the vector sum of all forces acting on the object. (Also called the net force, Fnet or ΣF). • Forces add together like vectors according to the rules we’ve learned previously. • We can use Pythagorean theorem and Law of Cosines to add forces at angles. • We can find force components, Fx and Fy using trig ratios.

9. Pulley Example • Muscleman is trying to lift a piano (slowly) up to a second story apartment. He is using a rope looped over two pulleys as shown in Fig 4-24. How much of the piano’s 2000N weight does he have to pull on the rope?

10. Pulley Solution • Looking at the forces on the lower pulley, we have 2 Tension force sections opposing the weight of the piano. Since the piano moves up slowly (a = 0m/s/s), then using Newton’s second law: 2FT –mg = ma. • To move the piano requires a tension in the cord of FT = mg/2. Muscleman exerts a force equal to half the piano’s weight. • We say this pulley has given a mechanical advantage of 2, since we exert 2x less force.

11. Example 4-13: Vector components at angles • Look at this example on page 95 and follow through the solution.

12. Your turn to Practice • Please do Chapter 4 Review pgs 104-105 #s 6, 7, 8, 10, 13, 16, 18, and 28.