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Newton’s Laws and Friction
• The friction model
– Experimental observations
– Theoretical models
– Coefficient of friction model (simple)
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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