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Newton’s Laws and Friction • The friction model – Experimental observations – Theoretical models – Coefficient of friction model (simple) • • • • Friction example 4-16 Pulling 10 kg box 4-19 Block on inclined plane with friction 4-21 Block on table, over side with friction 4-20 Coefficient of Friction • Good empirical “model”. Real science too complicated! • Basic observations: – Friction force varies between zero and some upper limit (at rest) – Friction force varies with normal force (not weight) – Friction force varies with surface characteristics – Friction force less when moving • Fs ≤ μs FN (static) • Fk = μk FN (kinetic) Coefficient of Friction Example 4-16 • 10 kg box, μs = 0.4 , μk = 0.3 • Determine Ffriction for applied force: – 0N – 10 N – 20 N – 38 N – 40 N • Determine acceleration at 40 N Block on Table with Friction (4-19) • Free Body Diagram • 𝐹𝑃 − 40𝑁, 𝜇 = 0.3 • Y direction ΣF = 0 – 𝐹𝑁 + 𝐹𝑃 sin 𝜃 − 𝑚𝑔 = 0 – 𝐹𝑁 = 78𝑁 • X direction ΣF = ma – 𝐹𝑃 cos(𝜃) − 𝐹𝐹𝑟 = 𝑚 𝑎𝑥 – 34.6𝑁 − 23.4 𝑁 = 10 𝑘𝑔 𝑎𝑥 – 𝑎𝑥 = 1.1 𝑚 𝑠 2 • Note: friction must be less then FP cos(ϴ) to move! Block on Incline with Friction (4-21) • FBD for block • Setup equations in y’ and x’ • Tilted y’ – Balances mg vs. normal force • Tilted x’ – Balances mg vs. friction, with leftover ma • Solve equations for acceleration ax • Find velocity • Static case – when does block start to move? Example 4-21 • 2nd Law in y’ and x’ – Tilted y’ 𝐹𝑁 − 𝑚𝑔𝑐𝑜𝑠𝜃 = 0 – Titled x’ 𝑚𝑔𝑠𝑖𝑛𝜃 − 𝐹𝐹𝑟𝑖𝑐 = 𝑚𝑎 • Solution – Combine x & y 𝑚𝑔𝑠𝑖𝑛𝜃 − 𝜇𝑘 𝑚𝑔𝑐𝑜𝑠𝜃 = 𝑚𝑎 – Rearrange 𝑎 = 𝑔𝑠𝑖𝑛𝜃 − 𝜇𝑘 𝑔𝑐𝑜𝑠𝜃 – Note m cancels out! – Static case 0 = 𝑔𝑠𝑖𝑛𝜃 − 𝜇𝑘 𝑔𝑐𝑜𝑠𝜃 𝜇𝑘 = 𝑡𝑎𝑛𝜃 One block on table, one over side (4-20) • FBD diagram for each • Y problem for A, normal force • X problem for A, friction force • Y problem for B, prevailing direction • Set up simultaneous equations • Solve • Limiting cases, Ffr = 0, ma = 0 Example 4-20 • Equations – Mass A vertical 𝐹𝑁 − 𝑚𝐴 𝑔 = 0 – Mass A horizontal 𝐹𝑇 − 𝐹𝐹𝑟𝑖𝑐 = 𝑚𝐴 𝑎 – Mass B vertical 𝑚𝐵 𝑔 − 𝐹𝑇 = 𝑚𝐵 𝑎 – Friction 𝐹𝐹𝑟𝑖𝑐 = 𝜇𝑘 𝐹𝑁 = 𝜇𝑘 𝑚𝐴 𝑔 • Note positive down for B – Solving A & B 𝑚𝐵 𝑔 − 𝜇𝑘 𝑚𝐴 𝑔 = (𝑚𝐵 +𝑚𝐴 )𝑎 – Solution 𝑎= 𝑚𝐵 𝑔−𝜇𝑘 𝑚𝐴 𝑔 𝑚𝐵 +𝑚𝐴 – Limiting case 𝜇𝑘 = 0, 𝑚𝐴 = 0 – Find FT Review • Other examples? – Prob 23 (tightrope walker) – Prob 49 (crate on flatbed truck) – Prob 32 (fuzzy dice) – Prob 26 (lawnmower) – Prob 25 (2 buckets) – Prob 45 (bobsled) – Prob 48 (2 crates) – Prob 10,13,15 (elevator) Exam 1 – Remember! Marine Base Kane‘ohe Bay, HI - June 2009 desert battle simulator • For projectile motion: – ay = -9.8 m/s2 – ax = 0