Why is it that tension is constant in a rope with negligible mass and frictionless pulley even if the weights are different?

Answer 1

The tension is only the same with the assumptions that the mass of the rope is negligible and that the pulley is frictionless. In real life situations, the tension will actually be different. The above assumptions are made to make calculations easier. To see why, let's set up some equations.

Effects of massless rope:

The sum of forces acting on the rope itself will be:

F_net = m_ropea_rope = T₁ - T₂ - m_ropeg

Solving for T₂, we get:

T₂ = T₁ - m_ropea_rope - m_ropeg

As long as the rope has mass, T₁ and T₂ are unequal and will be different.

Now if we assume that the rope is massless, the F_net equation simplifies to:

0 = T₁ - T₂ → T₂ = T₁

Effects of frictionless pulley:

Normally, pulleys experience a force called belt friction, which resists motion. The relative tensions due to belt friction are determined by the following equation:

T₂ = T₁e^μk*θ (μ_k is the coefficient of friction and θ is the angle of contact between the rope and the pulley)

As long as the pulley has friction, T₁ and T₂ are unequal and will be different.

Now if we assume that the pulley is frictionless, we can set the coefficient of friction to 0 and the equation simplifies to:

T₂ = T₁e⁰ → T₂ = T₁

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I hope this helped clarify! Please let me know if you have any additional questions.

Answer 2

I assume that you are thinking of the Atwood's machine geometry. If the pulley has some mass, then the tension in the part of the rope on the right is different from that in the part of the rope on the left.

This leads to a net torque on the pulley. That is OK, because if the pulley has mass, it will have a nonzero moment of inertia (rotational inertia). This leads to a reasonable angular acceleration. However, if the pulley has no mass, then a net torque would lead to an infinite angular acceleration!  It can be concluded that the net torque must be very nearly zero if the pulley has no mass. This, in turn, leads to the conclusion that the two tensions must be equal. This argument can be made more rigorous by a full free body diagram involving both masses and the pulley.