two blocks 3 kg and 2 kg are suspended from a rigid support by two inextensible wires each of length 1m and having linear mass density 0.2 kg/m . find the tension at the midpoint of each wire as the arrangementgets un upward acceleration of 2 m/s2.
The question above gives two possible cases.
Case: 1
Here length of each string L = 1 m
Mass per unit length (m/L) = 0.2 kg/m
As the system is moved upwards with an acceleration a = 2ms-2 Mass of the string below O = 0.2×0.5 = 0.1 kg. Let T be the tension at O.
Total mass of the system below point O (m’) = (2 + 0.1) kg = 2.1 kg
By FBD we have
T-m’g = m’a
T = m’(a + g)
=>T = 2.1×(10 + 2)
=>T = 25.2 N
Similarly,
Now tension experienced by the point O downwards in mass of 3 kg
T = (3+0.1)×(10 + 2) = 37.2 N
Case: 2
Let T be the tension at the point O of the 1st string
In case of 1st string the total mass m' below point O is = (0.1 + 3 + 0.2 + 2) kg = 5.3 kg
By FBD we have
T- m'g = m' a
=>T = m' (a + g)
=>T = 5.3×(10 + 2) = 63.6 N
Again let T’ be the tension at the middle point of 2nd string
Now mass of the system below the middle point of the 2nd string (m') = (0.1 + 2) = 2.1 kg
By FBD we have
T'- m'g= m'a
=> T'= m'(a + g)
=> T'= 2.1×(2 + 10) = 25.2 N