Question

Two blocks 3 kg and 2 kg are suspended from a rigid support by two inextensible wires, each of
length 1 m and having linear mass density 0.2 kg/m. find the tension at the mid- point of each wire
as the arrangement gets an upward acceleration of 2 m/s2 ?​

Answer 1

Consider the wire connecting the block of mass 3 kg with the ceiling. Let its tension be T₁.

The tension T₁ is acting upward at the midpoint of the wire.

The mass below the midpoint of the wire is,

m₁ = [(0.2 × 0.5) + 3 + (0.2 × 1) + 2] kg

m₁ = [0.1 + 3 + 0.2 + 2] kg

m₁ = 5.3 kg

As the system gets an upward acceleration a = 2 m/s²,

T₁ - m₁g = m₁a

T₁ = m₁ (a + g)

T₁ = 5.3 (2 + 10)

T₁ = 63.6 N

Consider the wire connecting the two blocks. Let its tension be T₂.

The tension T₂ is acting upward at the midpoint of this wire.

The mass below the midpoint of the wire is,

m₂ = [(0.2 × 0.5) + 2] kg

m₂ = [0.1 + 2] kg

m₂ = 2.1 kg

As the system gets an upward acceleration a = 2 m/s²,

T₂ - m₂g = m₂a

T₂ = m₂ (a + g)

T₂ = 2.1 (2 + 10)

T₂ = 25.2 N

Hence the tension in the upper string is 63.6 N and that in the lower string is 25.2 N.

Answer 2

Solution :-

• M 1 = 3 kg
• M 2 = 2 kg
• Length of each wire(l) = 1 m
• Linear mass density of wire(n) = 0.2 kg /m
• Acceleration(a) = 2 m/s ²
• Acceleration due to gravity (g) = 10 m/s ²

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First let us find out the tension at the midpoint of lower wire

Net tension acting on the block

• T = M ( g + a )

⟹ Net mass = Mass of the block + mass of the wire

⟹ Net mass = M 2 + l /2 x 0.2

⟹  Net mass = 2 + 1 /2 x 0.2

⟹ Net mass = 2 + 0.5 x 0.2

⟹ Net mass = 2 + 0.1

⟹ Net mass = 2.1 kg

⟹ Net acceleration = 10 + 2 = 12 m /s ²

Now let's substitute the values in the above equation ,

⟹ T = 2.1 x 12

⟹ T = 25.2 N

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Now let us find out the tension at the midpoint of the upper wire

Net tension acting on the block

• T '= M '( g + a )

⟹ Net mass = Mass of the block + mass of the wire

⟹ Net mass =(  M 1 + M 2 ) + mass of 0.5 m wire + mass of 1 m wire

⟹ Net mass = 5  + (0.5 x 0.2 + 0.2 x 1 )

⟹ Net mass = 5 +( 0.1 + 0.2)

⟹ Net mass = 5.3 kg

⟹ Net acceleration = 10 + 2 = 12 m /s ²

Now let's substitute the values in the above equation ,

⟹ T' = 5.3 x 12

⟹ T' = 63.6 N