Question

a ladder is hanging from ceiling three men A B and C of masses 40 kg 60 kg and 50 kg respectively are climbing the ladder. man A is going with retardation 2 metre per second square, B is going up with an acceleration of 1 metre per second square and man C is going with constant speed of 0.5 metre per second. Find the tension in string supporting the ladder?

Answer 1

Answer:

1450 N

Explanation:

As shown in figure, let A, B, C are moving with accelerations a₁, a₂ and a₃. If forces acting on them are F₁, F₂ and F₃ respectively then the tension T in the string will be equal to sum of all the forces i.e.

T = F₁ + F₂ + F₃

Now we need to find F₁, F₂ and F₃

Let the masses of A, B, C be m₁, m₂ and m₃ respectively

Then,

m₁ = 40 kg

m₂ = 60 kg

m₃ = 50 kg

And

a₁ = -2 m/s²

a₂ = 1 m/s²

a₃ = 0

For person A from Newton's second law of motion

F₁ - m₁g = m₁a

F₁ - 40g = 40 x (-2)

F₁ = -80 + 40 x 9.8 = 312 N

Similarly for B

F₂ - m₂g = m₂a

F₂ - 60 x 9.8 = 60 x 1

F₂ = 648 N

And for C

F₃ - 50g = 50 x 0

F₃ = 50 x 9.8 = 490 N

∴ Tension in the string T = 312 + 648 + 490 = 1450 N

Answer 2

Here is your answer.

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