1 / 15

Applications & Examples of Newton’s Laws

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:

Antony
Download Presentation

Applications & Examples of Newton’s Laws

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Applications & Examples of Newton’s Laws

  2. 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)

  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.

  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: ∑Fx = max ∑Fy = may Notice that this is theLASTstep, NOTthe first!

  5. Conceptual Example Moving at constant v, withNO friction, which free body diagram is correct?

  6. 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

  7. 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

  8. 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!!

  9. 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

  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 5.10Inclined Plane, 2 Connected Objects

More Related