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, F N.

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, F N on an object is not cancelled by Earth’s F g 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, F N = ?  F g = mg = 10.0 kg * 9.80m/s/s = 98 N acting downward. The only other force on the box is the normal force, F N and acts upward.  Since the box is at rest and the upward direction is positive, the sum of the y forces =0 so F N = 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, ΣF y =0, so F N -mg-40.0N = 0 and F N = 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 F p = 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, F net 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, F x and F y 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: 2F T –mg = ma.  To move the piano requires a tension in the cord of F T = 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.