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 45^o 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
F_f and from fig (a) with the drag force of 45lb.
S F_n = ma_n
4 cos45^o = (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 U_1-2 = T₂
1/2(4)[p(3) - l_o]² - 1/2(4)[3p/4(3) - l_o]² - 4(3sin45^o) = 1/2(4/32.2) (5.844)²
l_o = 5.51 ft

So the unstretched length of the cord so that the crate begins to leave the semicylinder at the instant of 45^o is l_o = 5.51 ft