Answer:
(a) 218.6 N
(b) 97.14 N
Step-by-step explanation:
When the system is in equilibrium, the net torque on the system is zero.
AC = 1.5 m, CD = 2.3 m, DB = 5 - 1.5 - 2.3 = 1.2 m
Let the centre of gravity of plank is at G.
AG = 2.5 m, CG = 2.5 - 1.5 = 1 m, GB = 2.5 m
(a) Let the reaction at C is R and at D is R'.
R + R' = 29 x 9.8 = 284.2 N ... (1)
Take the torque about C.
29 x 9.8 x CG = R' x GD
29 x 9.8 x 1 = R' x 1.3
R' = 218.6 N
(b) Take the torque about D.
6 x 9.8 x AD = R x CD
6 x 9.8 x (1.5 + 2.3) = R x 2.3
R = 97.14 N
Answer is B: 150
write two equations
r+s=190
4r+3s=720
solve by substitution
the get r=150 s=40
Where is the rest of the question?
