Applications Examples of Newtons Laws Forces are VECTORS

Applications & Examples of Newton’s Laws

• 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)

• 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.

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

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

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

Example Particle Under a Net Force Example (a) Crate being pulled to right across a floor. (b) Free body diagram for crate. ∑Fx = T = max ax = (T/m) ay = 0, because of no vertical motion. ∑Fy = 0 n – Fg = 0; n = Fg = mg

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. ay = 0, because of no vertical motion (equilibrium). ∑Fy = 0 n – Fg - F = 0 n = Fg + F = mg + F Showing again that the normal force is not always = & opposite to the weight!!

Example 5. 4: Traffic Light at Equilibrium (a) Traffic Light, Fg = 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. ay = 0, no vertical motion. ∑Fy = 0 T 3 – Fg = 0 T 3 = Fg = mg = 122 N (c) Free body diagram for cable junction (zero mass). T 1 x = -T 1 cos(37°), T 1 y = T 1 sin(37°) T 2 x = T 2 cos(53°), T 2 y = T 2 sin(53°), ax = ay = 0. Unknowns are T 1 & T 2. ∑Fx = 0 T 1 x + T 2 x = 0 or -T 1 cos(37°) + T 2 cos(53°) = 0 (1) ∑Fy = 0 T 1 y + T 2 y – 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

Example

Example 5. 6: Runaway Car

Example 5. 7: One Block Pushes Another

Example 5. 8: Weighing a Fish in an Elevator

Example 5. 9: Atwood Machine

Example 5. 10 Inclined Plane, 2 Connected Objects