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## Applications & Examples of Newton’s Laws

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**Forces are VECTORS!!**• Newton’s 2nd Law: ∑F = ma ∑F = VECTORSUMof all forces on mass m Need VECTORaddition to add forces in the 2nd Law! • Forces add according to rules of VECTOR ADDITION! (Ch. 3)**Newton’s 2nd Law problems:**• STEP 1:Sketch the situation!! • Draw a “Free Body” diagram for EACHbody 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: ∑Fx = max ∑Fy = may Notice that this is theLASTstep, NOTthe first!**Conceptual Example**Moving at constant v, withNO friction, which free body diagram is correct?**ExampleParticle 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**ExampleParticle 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**ExampleNormal 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 T3 – Fg = 0 T3 = Fg = mg = 122 N (c) Free body diagram for cable junction (zero mass).T1x = -T1cos(37°), T1y = T1sin(37°) T2x = T2cos(53°), T2y = T2sin(53°), ax= ay= 0. Unknowns are T1 & T2. ∑Fx = 0 T1x + T2x = 0 or -T1cos(37°) + T2cos(53°) = 0 (1) ∑Fy = 0 T1y + T2y – T3 = 0 orT1sin(37°) + T2sin(53°) – 122 N = 0 (2) (1) & (2) are 2 equations, 2 unknowns.Algebrais required to solve for T1& T2! Solution: T1 = 73.4 N, T2 = 97.4 N