Question

Three blocks A,B and C of masses 3 kg,2 kg and 7 kg respectively are connected as shown.Determine the acceleration of A,B and C.Also find the tension in the string.

Given : m_A=3kg
           m_B=2 kg
           m_C=7kg 
To find: Acceleration of blocks A,B and C

 

Answer 1

Assuming the pulleys and the connecting inextensible strings are massless and frictionless
Assume xA,xB,xC and xD be the displacements of blocks A,B,C and D respectively.

Assume blocks A,B,C and D move downwards.So xA,xB,xC and xD will increase

Assume k be the length of string that remains constant irrespective of positions of A,B and C.
As the length of string is constant
(xA-xD)+(xB-xD)+k=0
xA+xB-2xD+k=0
Differentiating w.r.t t
v_A+v_B-2v_D=0
Differentiating once again w.r.t to t
aA+aB-2aD=0 …….(1)
and xD+xC+k=0
xD=-xC-k
Differentiating w.r.t t

vD=-vC
Differentiating once again w.r.t to t
aD=-aC ………..(2)
Substituting (2) in (1)
aA+aB+2aC=0 …..(3)
Assume tensions T1 and T2 be the tensions in two strings
For block A

ΣF=mAaA

3g-T1= mAaA
T1=3g-3aA ……….(4)
For block B

ΣF=mBaB
2g-T1=mBaB
2g-(3g-3aA)=2aB (From 4)
3aA-2aB=g ……….(5)
For pulley D

ΣF=mBaB
`2T_1-T_2=m_Da_D`
`m_D`=0
`2T_1-T_2=0`
`T_2=2T_1`
     `=2(3_g-3a_A)`
     `=6g-6a_A` ………(6) (From 6)
For block C

`ΣF=m_Ba_B`

`7_g-T_2=m_Ca_C`

`7_g-(6_g-6a_A)=7aC` ………..(From 6)

`6_aA-7_aC=-g` ………..(7)

Solving (3),(5) and (7)

`a_A=0.4988 m/s^2`

`a_B=-4.1568 m/s^2`

`a_C=1.8290 m/s^2`

From (4)

`T_1=3_g-3a_A`

       =3(9.81-0.4988)

       =27.9336 N

From (6)

`T_2=2T_1`

=55.8671 N

Acceleration of block A=0.4988 m/s²(Vertically downwards) Acceleration of block B=4.1568 m/s²(Vertically upwards) Acceleration of block C=1.8290 m/s²(Vertically downwards)
Tension of the string T1=27.9336 N