1. Applications & Examples of Newton’s Laws

2. Forces are VECTORS!! Newton’s 2 nd Law: ∑F = ma ∑F = VECTOR SUM of all forces on mass m  Need VECTOR addition to add forces in the 2 nd Law! –Forces add according to rules of VECTOR ADDITION! (Ch. 3)

3. Newton’s 2 nd Law problems: STEP 1: Sketch the situation!! –Draw a “Free Body” diagram for EACH body in problem & draw ALL forces acting on it. Part of your grade on exam & quiz problems! STEP 2: Resolve the forces on each body into components –Use a convenient choice of x,y axes Use the rules for finding vector components from Ch. 3.

4. STEP 3: Apply Newton’s 2nd Law to EACH BODY SEPARATELY : ∑F = ma –A SEPARATE equation like this for each body! –Resolved into components: ∑F x = ma x ∑F y = ma y Notice that this is the LAST step, NOT the first!

5. Conceptual Example Moving at constant v, with NO friction, which free body diagram is correct?

6. Example Particle in Equilibrium “Equilibrium” ≡ The total force is zero. ∑F = 0 or ∑F x = 0 & ∑F y = 0 Example (a) Hanging lamp (massless chain). (b) Free body diagram for lamp. ∑F y = 0  T – F g = 0; T = F g = mg (c) Free body diagram for chain. ∑F y = 0  T – T´ = 0; T ´ = T = mg

7. Example Particle Under a Net Force Example (a) Crate being pulled to right across a floor. (b) Free body diagram for crate. ∑F x = T = ma x  a x = (T/m) a y = 0, because of no vertical motion. ∑F y = 0  n – F g = 0; n = F g = mg

8. Example Normal Force Again “Normal Force” ≡ When a mass is in contact with a surface, the Normal Force n = force perpendicular to (normal to) the surface acting on the mass. Example Book on a table. Hand pushing down. Book free body diagram.  a y = 0, because of no vertical motion (equilibrium). ∑F y = 0  n – F g - F = 0  n = F g + F = mg + F Showing again that the normal force is not always = & opposite to the weight!!

9. Example 5.4 : Traffic Light at Equilibrium (a) Traffic Light, F g = mg = 122 N hangs from a cable, fastened to a support. Upper cables are weaker than vertical one. Will break if tension exceeds 100 N. Does light fall or stay hanging? (b) Free body diagram for light. a y = 0, no vertical motion. ∑F y = 0  T 3 – F g = 0 T 3 = F g = mg = 122 N (c) Free body diagram for cable junction (zero mass). T 1x = -T 1 cos(37°), T 1y = T 1 sin(37°) T 2x = T 2 cos(53°), T 2y = T 2 sin(53°), a x = a y = 0. Unknowns are T 1 & T 2. ∑F x = 0  T 1x + T 2x = 0 or -T 1 cos(37°) + T 2 cos(53°) = 0 (1) ∑F y = 0  T 1y + T 2y – T 3 = 0 or T 1 sin(37°) + T 2 sin(53°) – 122 N = 0 (2) (1) & (2) are 2 equations, 2 unknowns. Algebra is required to solve for T 1 & T 2 ! Solution: T 1 = 73.4 N, T 2 = 97.4 N

10. Example

11. Example 5.6: Runaway Car

12. Example 5.7: One Block Pushes Another

13. Example 5.8: Weighing a Fish in an Elevator

14. Example 5.9: Atwood Machine

15. Example 5.10 Inclined Plane, 2 Connected Objects