Question 1)
Rod AB of length “l” and negligible weight lies in a vertical plane and is pinned to blocks A and B. The weight of each block is W, and the blocks are connected by an elastic cord which passes over a pulley at C. Neglecting friction between the blocks and the guides, determine the value of θ as a function of the tension in the cord, T (known).
Using cos(a+b) = cos(a)cos(b) – sin(a)sin(b)
sin(a+b) = sin(a)cos(b) + cos(b)sin(a)
W*(cos(θ)cos(60) – sin(θ)sin(60)) + T*(sin(θ)cos(30) + cos(θ)sin(30) – cos(θ)cos(30) + sin(θ)sin(30)) = 0

 ­W*tan(θ)sin(60) + T*tan(θ)*(cos(30) + sin(30)) + W*cos(60) + T*(sin(30) – cos(30)) = 0
 tan(θ)(­W*sin(60) + T*(cos(30)+sin(30)) = T*(cos(30) – sin(30)) – W*cos(60)
 θ = tan­1(T*(cos(30) – sin(30)) – W*cos(60)) = tan­1(T*(1 – sqrt(3)) – W)
­W*sin(60) + T*(cos(30)+sin(30)) (­W*sqrt(3) – T*(sqrt(3) +1) Question 2) Knowing that w=300 N/m, determine a) The resultant concentrated load on the beam and its location.
b) The smallest distance a for which the vertical reaction at support B is 2.5 times that at A. The resultant concentrated load has magnitude 2400 + 1200(8­a)
 Rtot = 1200(10­a)
The location of the resultant can be found by letting z be the horizontal distance from R1:
R1*z = R2 * ((4 – (8­a)/3) ­ z)
So z = R2*(4­(8­a)/3) = 1200(8­a)*(12 – (8­a)) = (8­a)*(12 – (8 – a)) = 32 +4a ­ a
2
(R1 + R2) 1200*3*(10­a) 3 * (10 – a) 3*(10­a)
Note that 4 + z is the horizontal distance from A => total distance = a 2 + 8a ­152 3*(a – 10)
b) To find a where By = 2.5*Ay, just set them equal:
(2.5)*(1200 + 50*(8­a)2) = 1200 + 50(16+a)*(8­a)
 a = 3.36m, 5.78m (3.36 is smaller)