Question

Three Equal weights of mass 2 kg each are hanging on a string passing over a fixed pulley as shown in fig. What is the tension in the string connecting weights B and C ?

◆ Options Are Given ◆

♀ Answer : 13.3 N . How ?

Answer 1

Hola,

Let us take the tension between A and B be T and tension between B and C be T1.
The masses are connected to a massless string moving over a frictionless pulley.
Now, if they are allowed to move freely, then they move with common magnitude of acceleration 'a'.
Mass of A, B and C = 2kg (m)

Now,

For C,
=> ma = mg - T1

=> T1 = mg - ma. ....(1)

For B,
=> ma = mg + T1 - T. ....(2)

For A,
=> ma = T - mg

=> T = ma + mg. ....(3)

Now, from (1) and (3) we will put value of The and T1 in (2), we get

=> ma = mg + (mg - ma) - (ma + mg)

=> ma = mg + mg - ma - ma - mg

=> 3ma = mg

=> a = g/3

Now, putting the value in (1) and (3), we will get the tensions..

Tension between B and C = m(g - a)

= m(g - g/3)

= 2(2g/3)

= 4g/3. (g = 10)

= 40/3

= 13.3N

Tension between A and B = m(a + g)

= m(g/3 + g)

= 2(4g/3)

= 8g/3

= 80/3

= 26.6N

From above we get that the tension in the string connecting weights B and C = 13.3N (Option B)

Hope this helps....:)

Answer 2

Let the required tension be T and the tension in the string connecting A and B be T 1 

equations are: 

for C = 2 g - T = 2 a            (i) 

for B = 2 g + T - T 1 = 2 a           (ii) 

and T 1 - 2 g = 2 a        (iii) 

adding (i), (ii), (iii) 

2 g = 6 a or a = g/3 

substituting in (i) = T = 2(g -a) = 2(g - g/3) = 4 g/3 

= 4(9.81)/3 = 13.08 N