It's about mechanics of materials quiz QnA, Quizzes of Statistical mechanics

Download It's about mechanics of materials quiz QnA and more Quizzes Statistical mechanics in PDF only on Docsity! ∑ ∑ 2.001 - MECHANICS AND MATERIALS I Lecture #4 9/18/2006 Prof. Carol Livermore TOPIC: FRICTION EXAMPLE: Box on floor μs =Coefficient of Static Friction FBD Equation of equilibrium Fy = 0 N − W = 0 N = W Fx = 0 1 T − F = 0 T = F At impending motion only : F = μsN For well lubricated, μs ≈ 0.05. For very clean surfaces μs ≈ 0.4 − 1. After it starts to move: F = μkN μk = Coefficient of kinetic friction. μk < μs EXAMPLE: Block on an inclined plane Q: At what angle (α) does the block slide down the plane? FBD: 2 Case 2: Impending motion is up the plane. ∑ Fx = 0 T2 + F2 − W sin α = 0 ∑ Fy = 0 N2 − cos α = 0 What about T? FBD of Cable Look at differential element 5 ( ) ( ) ( ) ( ) Fx = 0 dθ dθ T (θ) cos −T (θ + dθ) cos = 0 2 2 T (θ) = T (θ + dθ) = T Fy = 0 dθ dθ dN − T (θ) sin −T (θ + dθ) sin = 0 2 2 dθ dθ dN − T (θ) − T (θ + dθ) = 0 2 2 Tdθ = dN So: T = W0 Back to block: T1 = T2 = W0 N1 = N2 = N = W cos α For case 1: F1 = μsN = μsW cos α μsW cos α + W0 + W sin α = 0 W0 = W sin α − μsW cos α The block will be stable against downward motion when: W0 = W sin α − μsW cos α For case 2: F2 = μsN = μsW cos α μsW cos α + W0 + W sin α = 0 W0 = W sin α − μsW cos α 6 ∑ ( ) ( ) ( ) ( ) The block will be stable against downward motion when: W0 ≤ W sin α + μsW cos α So it is stable when: W (sinα − μs cos α) ≤ W0 ≤ W (sinα + μs cos α) What about pulley with friction? Look at a differential element. Recall a rope around a rod. Fx = 0 T (θ) cos dθ 2 −T (θ + dθ) cos dθ 2 −dF = 0 ∑ Fy = 0 ( ) ( ) dθ dθ dN − T (θ) sin −T (θ + dθ) sin −dF = 0 2 2 dθ dθ sin ≈ 2 2 dθ cos ≈ 1 2 dT = T (θ + dθ) − T (θ) ⇒ T (θ + dθ) = T (θ) + dT = T + dT So: T (θ) − T (θ + dθ) − dF = 0 dT = −dF Tdθ dθ dN − + (T + dT ) = 0 2 2 T + dT → 0 dN − Tdθ = 0 With impending motion: dF = μsdN dT = −μsdN