Answer:FCF = 14.8998 N (Compression)
FBE = 52.7342 N (Compression)
Explanation:We have to find the mass center of ABCD which is at G.
XCG = (0.45 m /2) = 0.225 m
YCG = ∑mi*yi / ∑mi
⇒ YCG = (3 Kg*0.225m+3 Kg*0.225m+3 Kg*0m) / (3*3Kg) = 0.15 m
Now, we can apply ∑F = m*a
where a is the tangential acceleration of the links
M = 3*m = 3*3 Kg = 9 Kg
∑F = m*a ⇒ M*g*Sin 40º = M*a ⇒ a = g*Sin 40º
⇒ a = (9.81 m/s²)*Sin 40º = 6.3057 m/s²
We apply ∑MB as follows
∑MB = (0.45)*(FCF*Sin 50º) - M*g*(0.225) = - (M*a*Sin 40º)*(0.225) - (M*a*Cos 40º)*(0.15)
⇒ (0.45)*(FCF*Sin 50º) - 9*9.81*(0.225) = - (9*6.3057*Sin 40º)*(0.225) - (9*6.3057*Cos 40º)*(0.15)
⇒ FCF = 14.8998 N (Compression)
Then
∑Fy = m*ay
⇒ FCF*Sin 50º + FBE*Sin 50º - M*g = - M*a*Sin 40º
⇒ 14.8998*Sin 50º + FBE*Sin 50º - 9*9.81 = - 9*6.3057*Sin 40º
⇒ FBE = 52.7342 N (Compression)
We can see the system in the pic shown.
