Shows an atwood machine that consists of two blocks (of masses m1 and m2) tied together with a massless rope that passes over a fixed, perfect (massless and frictionless) pulley. in this problem you'll investigate some special cases where physical variables describing the atwood machine take on limiting values. often, examining special cases will simplify a problem, so that the solution may be found from inspection or from the results of a problem you've already seen. for all parts of this problem, take upward to be the positive direction and take the gravitational constant, g, to be positive.
part a consider the case where m1 and m2 are both nonzero, and m2> m1. let t1 be the magnitude of the tension in the rope connected to the block of mass m1, and let t2 be the magnitude of the tension in the rope connected to the block of mass m2. which of the following statements is true?
consider the case where and are both nonzero, and . let be the magnitude of the tension in the rope connected to the block of mass , and let be the magnitude of the tension in the rope connected to the block of mass . which of the following statements is true?
a) t1 is always equal to t2.
b) t2 is greater than t1 by an amount independent of velocity.
c) t2 is greater than t1 but the difference decreases as the blocks increase in velocity.
d) there is not enough information to determine the relationship between t1 and t2.

Newton's second law allows us to find which result is correct for the question about the stresses in Atwood's machine is:

      a)  t1 is always equal to t2

Newton's second law gives the relationship between the net force, the mass and the acceleration of the body

        ∑ F = m a

where F is force, m is mass and acceleration.

A free-body diagram is a diagram of the forces without the details of the bodies, in the attached we see a free-body diagram of an Atwood machine.

Let's write Newton's second law for each field

Body 1

            T₁ - W₁ = m₁ a

Body 2

            W₂ -T₂ = m₂ a

The rope to keep the stretched, the acceleration of the two bodies is the same and since the pulley is ideal, it has no mass, the tension of the two rope is the same.

The system solution is:

           (m₂ - m₁) g = (m₁ + m₂) a

Therefore we see that the tension of the rope is the same.

Let's review the claims:

a) True. As the pulley is ideal the tensions are equal.

b) False. The tension depends on the acceleration.

c) False. In an ideal pulley the tensions are equal.

d) False. With the data indicated, the problem can be completely solved.

In conclusion, using Newton's second law we can shorten which result is correct for the question about the stresses in Atwood's machine  is:

     a) t1 is always equal to t2.

Learn more about Newton's second law here: link

The correct answer to the question is

At equilibrium a) t1 is always equal to t2.

Explanation:

For a Atwood machine we have the masses m₁ and m₂ tied together by a string

Where m₂ > m₁ we take upward to be the positive direction and gravitational constant g = + ve

When in equilibrium,  by analyzing the tension in the string, we have

T₁, tension is due to the weight of m₁ and the reaction of m₂

similarly for T₂ tension is due to the weight of m₂ and the reaction of m₁

Since the string is assumed to be weightless and continuous, and the pulley is friction-less,  the two weights is therefore supported only by the string hence the   tension T₁ and T₂ are equal

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