An elastic cord having a stiffness k = 4lb/ft is attached to the crate as shown and to the base of the semicylinder at point C.If the block is released from rest from the bottom of the cylinder . Determine the unstretched length of the cord so the crate begins to leave the semicylinder at the instant of 45o if the weight of the crate is 4lb. Neglect the size of the block. 1) 5.5 ft 2) 9.5 ft 3) 3.5 ft 4) 5.3 ft Equation of motion: Since the crate slides, the friction force developed between the crate and its surface is Ff and from fig (a) with the drag force of 45lb. S Fn = man 4 cos45o = (4/32.2)(v2/3) v2 = 34.15 (ft/s)2 v= 5.844 ft/s Since the crate is starting from rest there is no initial velocity also no energy so T1 = 0. Applying the conservation of energy principle we have: T1 + S U1-2 = T2 1/2(4)[p(3) - lo]2 - 1/2(4)[3p/4(3) - lo]2 - 4(3sin45o) = 1/2(4/32.2) (5.844)2 lo = 5.51 ft So the unstretched length of the cord so that the crate begins to leave the semicylinder at the instant of 45o is lo = 5.51 ft