Question

Three bars, each of mass 3 kg, are welded together and are pin-connected to two links BE and CF. Neglecting the weight of the links, determine the force in each link immediately after the system is released from rest.

Answer 1

FCF = 14.8998 N (Compression)

FBE = 52.7342 N (Compression)

Step-by-step 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.