Question

Consider the situation shown in the following figure. Calculate (a) the acceleration of the 1.0 kg blocks, (b) the tension in the string connecting the 1.0 kg blocks and (c) the tension in the string attached to 0.50 kg.

Answer 1

From the above diagrams:
T + ma − mg = 0
T + 0.5a − 0.5 g = 0                       (1)
μR + ma + T₁ − T = 0
μR + 1a + T₁ − T = 0                      (2)
μR + 1a − T₁ = 0
μR + a = T₁                                    (3)
From Equations (2) and (3) we have
μR + a = T − T₁
⇒ T − T₁ = T₁
⇒ T = 2T₁
So, Equation (2) becomes
μR + a + T₁ − 2T₁ = 0
⇒ μR + a − T₁ = 0
⇒ T₁ = μR + a
        = 0.2g + a                       (4)

and Equation (1) becomes
2T₁ + 0.5a − 0.5g = 0
`=> T_1=(0.5g-0.5a)/2`
= 0.25g - 0.25a                    (5)
From Equations (4) and (5)
0.2g + a = 0.25g − 0.25a
`=>a=0.05/1.25xx10`
= 0.4 x 10 m/s²                     [g = 10 m/s²]
Therefore,
(a) the acceleration of each 1 kg block is 0.4 m/s²,
(b) the tension in the string connecting the 1 kg blocks is
   T₁ = 0.2g + a + 0.4 = 2.4 N
​    and
(c) the tension in the string attached to the 0.5 kg block is 
    T = 0.5g − 0.5a
       = 0.5 × 10 − 0.5 × 0.4
       = 4.8 N.