Question

Block A has a mass of 10 kg, and blocks B and C have masses of 5 kg each. Knowing that the blocks are initially at rest and that B moves through 3 m in 2 s, determine (a) the magnitude of the force P, (b) the tension in the cord AD. Neglect the masses of the pulleys and axle friction.

Answer 1

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