Question

1. (20 Marks) An object has an L-shape and can be treated as 2 beams fixed together at right angle. Beam 1 (B) has a weight wi = 180 N and a length Li-2.00 m and Beam 2 (B) has a weight w2 90.0 N and a length L2- 1.00 m. This object is hinged by its corner O to the ground via a hinge that allows it to rotate clockwise counter clockwise, when possible. B2 rope B2 0 Figure 2 Figure 1 a) If the object is in equilibrium, as shown in figure 1 above, find the value of the angle 8. b) Show that the force from the hinge on the object is vertical and evaluate it. In figure 2, a block with a weight w - 40.0 N is attached (fixed) at the middle of Bi and a massless horizontal rope is attached at the top end of B2. e) Ife - 76", find the tension T in the rope that keeps the system in equilibrium d) Find the magnitude and direction of the force from the hinge on the object now.

Answer 1

Given length = 60 cm depth = 4.7 m Bottom surface will be at 5.3 m uses P_0 = 1.01 times 10^5 pa find difference in pressure we have delta p = rho g L = 10^3 times 10 times 0.6 delta rho = 6 times 10^3 pa (L = 60 cm = 0.6 m) average pressure P_avg = p_0 + rho g h/2 = 1.01 times 10^5 + (10^3 times 10 times 0.6/2) P_avg = 1.04 times 10^5 pa \r\n Evaluate mass & density we have m g = m g = V_in rho w g (from FBD) rightarrow m = V_in rho w = (0.6)^3 40/100 times 10^3 m = 86.4 kg now density rho = m/v = 86.4/(0.6)^3 rho = 400 kg/m^3 volume V_nollow = (0.6)^3 - 86.4/2400 V_nollow = 0.18 m^3 \r\n to give statement we have rho_avg = 86.4 + 1000 times 0.18/2 /(0.6)^3 rho_avg = 816.67 kg/m^3