1. Vb→ = W1. (BE)
Vb→ = Vd + Vd/b
W1.(BE) →= 1.33↓ + W2.(BD)
Va = Vb+Va/b
= 1.33↓ + W2.(BD) + W2.(AB)
Va = Vd + Va/d
= 1.33↓ + W2(AD)
1.33↓ + W2(AD) = 1.33↓ +
W2.(BD) + W2.(AB)
Analyze forces in x
direction:
W2.(AD) cos60 = W2.
(BD) cos60 + W2.(AB) cos60
W2.600/2 = W2.360/2 +
W2.240/2
W2 = 1 rad/s
So we can find W1
W1.(BE)→ = 1.33↓ + W2.(BD)
W1 = (W2.360.cos60)/198
W1 = 0.9 rad/s
Now Va:
Va = Vb+Va/b
= 1.33↓ + W2.(BD) + W2.(AB)
Analyze in direction x :
Va “x” = W2.(BD)cos 60 + W2.(AB) cos60 =0.3
Analyze in direction y :
Va”y” = W2.(BD)sin60 + W2.(AB)sin60 - 1.33 =0.5
Va = √ (Va”x”)2
+ (Va”y”)2
Va = 0.6 m/s
Tanθ = Va”y” / Va”x”
Θ = ≈60