Answer:
10 N
103.1 N
Explanation:
Solution:
Let the position coordinate y be positive downward.
Constraint of cord AD:
Ya+Yb=constant
Va+Vd=0
Aa+Ad=0
Constraint of cord BC:
(Yb-Yd)+(Yc-Yd)=constant
Vb+Vc-2Vd=0
Ab+Ac-2Ad=0
Eliminate Ad:
2Aa+Ab+Ac=0.....................................(1)
We have uniformly accelerated motion because all of the forces are constant.
Yb=Yb(o)+Vb(o)*t+1/2Ab*t^2 , Vb(o)=0
Ab=2[Yb-Yb(o)]/t^2=(2)(3)/(2)^2=1.5 m/s^2
Pulley D:
∑Fy=0: 2Tbc-Tad=0
Tad=2Tbc
Block A:
∑Fy=mAy: Wa-Tad=m_(a)Aa
Aa=Wa-Tad/m_(a)
=Wa-2Tbc/m_(a)............................(2)
Block C:
∑Fy=mAy: Wc-Tbc=m_(c)Ac
Ac=Wc-Tbc/m_(c)............................(3)
Substituting the value for Ab and Eqs. (2) and (3) into Eq. (1), and solving for Tbc
2(Wa-2Tbc/m_(a))+Ab+(Wc-Tbc=m_(c)Ac)=0
2(m_(a)g-2Tbc/m_(a))Ab+(m_(c)g-Tbc=m_(c)Ac)=0
(4/m_(a)+1/m_(c))*Tbc=3g+Ab
(4/10+1/5)*Tbc=3(9.8)+1.5
Tbc=51.55 N
Block B:
∑Fy=mAy: P+Wb-Tbc=m_(b)Ab
(a) Magnitude of P.
P=Tbc-Wb+m_(b)Ab
=51.55-5(9.81)+5(1.5)
=10 N
(b) Tension in cord AD.
Tad=2Tbc=(2)(51.55)=103.1 N
Note: find the attachment
