Question

The uniform crate has a mass of 50 kg and rests on the cart having an inclined surface. Determine the smallest acceleration that will cause the crate either tip or slip relative to the cart. What is the magnitude of this acceleration? The coefficient of static friction between the crate and the cart is mu_s = 0.5.

Answer 1

In order to get maximum acceleration that will make crate slip. Fx -F max =0.5N Consider the equilibrium of forces along the y-axis. Ncos15°+05Nsin15 50x9.81 N(cos15+0.5sin 15)-490.5 N 447.94 N Consider the equilibrium of forces along the x-axis F cos15-Nsin 15°-maG 0.5Ncos 15-sin 15ma 447.94 N (0.5cos 19-sin 15°) = 50aG 50aG-0.2241x 447.94 As = 2.008 m/s

Take moment about point A. -(mg cos159) (0.3)(mag cos15) (0.5) ) -( +mg sin 15°)(0.5 maç sin 15° (0.3 ) 9.81(0.3cos 15°-0.5sin 15°) 0.5cos150+0.3sin15 ag 2.81 m/s Thus acceleration of mass to slip in the first case is smaller than the acceleration required at tipping. So, slipping will occur first. Therefore, the smallest acceleration that will cause slipping or tipping is ag 2.008 m/s