Question

find the acceleration of the 6 kg block in the figure all the surface and Pulley are smooth also strings are inextensible and light. Take g=10 m/s^2

Answer 1

HELLO THERE!

By Free Body Diagram of the 2kg block hanging down,

T₁ - 2g = 2a

=> T₁ - 20 = 2a .... (i)

By free body diagram of the 2kg block placed on the inclined plane, we have:

T₂ - (2gsin 30 + T₁) = 2a

=> T₂ - 10 - T₁ = 2a

=> T₂ - T₁ = 2a + 10

=> T₂ = 2a + 10 + T₁ ..... (ii)

By free body diagram of the 6 kg block, we have:

6g - T₂ = 6a

=> 60 - 2a - 10 - T₁ = 6a

=> 50 - T₁ = 8a ....(iii)

(i): T₁ - 20 = 2a

(iii): 50 - T₁ = 8a

(i) + (iii) =

T₁ - 20 + 50 - T₁ = 8a + 2a

=> 30 = 10a

=> a = 3 m s⁻²

Hence, acceleration of the 6 kg block = 3 m s⁻².

Put a in Equation (i), to get the value of T₁ (Tension in pulley 1).

Put a and T₁ in Equation (ii), to get the value of T₂ (Tension in pulley 2).

HOPE MY ANSWER IS SATISFACTORY..

Thanks!

Answer 2

all the three blocks are attached with a single inextensible strings.

for 6kg block,
Mg - T = Ma
6 × g - T = 6 × a
6 × 10 - T = 6a
60 - T = 6a ............(i)

for inclined 2kg block,
T - mgsin30° - T' = ma
T' - 2 × 10 × 1/2 - T = 2a
T - 10 - T' = 2a.....................(ii)

for vertical 2kg block,
T' - m'g = m'a
T' - 2 × 10 = 2a
T' - 20 = 2a ..........(iii)

from equations (i) and (ii),
60 - 10 - T' = 6a + 2a
50 - T' = 8a ......(iv)

from equations (iii) and (iv),
30 = 10a
a = 30/10 = 3 m/s²

hence, acceleration of 6kg block is 3 m/s²