Title PSAD 3 - Friction.pdf ✅

PSAD 3: Friction Video Transcript Friction: f = μN ( relation between friction and normal force ) μ = tan ∅ ( coefficient and angle of friction ) Situation 1 – Refer to Fig. MECH0001 A 120 kg man walks on a 4 m plank resting on rough surfaces A and B. Given: Coefficient of friction at A = 0.40 Coefficient of friction at B = 0.20 1. At what distance x (m) will the plank start to slide? a. 2.33 c. 2.18 b. 1.67 d. 1.82 2. What is the friction force (N) at A? a. 197.17 c. 256.75 b. 98.59 d. 128.38 3. Find the vertical force (N) at B. a. 642 c. 691 b. 503 d. 493 Solution: Drawing the free body diagram, Using summation of components of forces, + →∑Fx = 0: NA cos 60° − fA cos 30° − fB = 0 NA cos 60° − (0.40NA ) cos 30° − (0.20NB) = 0 0.1536NA − 0.20NB = 0 eqn. (1) +↑ ∑Fy = 0: NA sin 60° + fA sin 30° + NB = 1177.2 N NA sin 60 + (0.40NA ) sin 30 + NB = 1177.2 N 1.066NA + NB = 1177.2 eqn. (2)

Solution: Drawing the free body diagram for impending sliding: Using summation of components of forces, + →∑Fx = 0: f − T cos 60° = 0 0.40N − T cos 60° = 0 T cos 60° − 0.40N = 0 eqn. (1) +↑ ∑Fy = 0: T sin 60° + N − 1324.35 = 0 T sin 60° + N = 1324.35 eqn. (2) Solving the equations simultaneously (Caltech: MODE – EQN – 2 unk): T = 625. 867 N N = 782.334 N Drawing the free body diagram for impending sliding, the normal force is at the edge of the log: Solving the tension: ↶ +∑MA = 0: T sin 60° (4) − 1324.35(2) = 0 T = 764. 614 N
The largest value that can be applied while maintaining equilibrium comes from the condition of sliding, Tsliding = 625. 867 N. Drawing the free body diagram for impending sliding, Solving for normal force N: +↑ ∑Fy = 0: N + T sin 60° = 1324.35 N + 625.867 sin 60 = 1324.35 N = 782. 334 N Solving for location: ↶ +∑MA = 0: T sin 60° (4) + N(2 − x) = 1324.35(2) 625.867 sin 60° (4) + 782.334(2 − x) = 1324.35(2) x = 1.389 m Finally, Location from A = 2 − x Location from A = 2 − 1.386 Location from A = 0. 614 m