Question

Q5. As shown in Fig. 5, a uniform sphere of mass m = 8 kg and radius r rolls without slip toward a large crate with mass M = 80 kg that rests on top of a rubber mat nailed into the ground. The coefficient of friction between the crate and the mat, 1 = 0.75. A spring with stiffness k is attached to the crate at a height h = 0.85m above the mat, and the crate's mass centre G is located a distance d = 0.55 m from either of its supports A and B. The sphere makes contact with the spring with a speed of v = 5 m/s. Determine the spring constant k so that the crate just barely tips without slipping, and explain briefly your answer. Draw appropriate free body, kinetic diagrams and reference system/s to clearly illustrate your logical solution steps. R M Ini

Answer 1

page Answer: - 6 K dissm to.sm Free body diagram: CoKg 819 Given m = 8kg, M = 80kg M = 0.75 V = 5 m/s Now, energy associated with 8kg mass E = 1 mort / I wr Isphere - 2 m and then, E = 1 mur + 1 / 2 ( 2 m ) . Imut I mor Tomor

page (2) = 1 x 8 x5 - 2 x8825 = 10x200 E = 140J Now, as the ball strikes, the energy associated with 8kg mass get store in the spring as spring PE Here & is deflection in spring a. I ka = 140J Ka- 280 x = 280 Now, the compressed spring will exert force F on the body M. F = ka - K., 280 N F = √280k N Now, when the crate just barely tips without Slipping, Taking moment about point B, we get CS Scanned with Fon - Mg.d + MM3.0 = 0 [closek wise positive

page3 - F. h = Mgd → V280K X Co.gs) – 80X9.81X0.55 = √280k * 0.85 = 431.64 = √280k = 431.64 0.85 431.64 k = 0.850 280 JK = 431.64 14.22305 = = UK = 30.3479 K = 920.995 N/m. . spring constant 'K' is 920.995 N/m