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 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!

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 ∑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

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

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!!

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

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