Question

1. A 50 lb uniform crate that is 1.5 ft wide is being pushed with a downward force directed at an angle of 30° from horizontal. The coefficient of static friction is 0.25. The crate has a flat bottom. a. What is the minimum force, P, required for impending motion of the crate to the right? Consider slipping and overturning (you should calculate 2 different values of P). b. Will the crate slip or overturn first? Why? - 11 ft 30° 3 ft

Answer 1

@ FBD of box supping I Case Pcossop 300 (30° Ing 31 AN. Psin3oo. få frictional force. N: Normal " under Equallaboiun. Στο - & Pros30° af 11 & N = Psin 30° + mg, roll # At Extseme condition, .. = MN f = 0.254 Pain 30 proto Put 0 in 0 Pcos 30o = 0.25 (p sin 30° +50). Px 8 = 0.25 (+50). BP = f +25. 1P = 16.86 lb. =

Case I overtun when block just FBD overturns. IH tw. N-O Takie moment about point o P op cos BO' Psin30°. I 0 W TO Pcos30°x 4 - P sin 300 (165) - WXIS) = 0. 8 49 20330° - P sim 30°x1.5 = 5OXIS o p [46930 ° - sin3031:57 - 37.5. p = 13.816 d.

D Since P in overtuig < Pin slipply. The block will overturn first