Respuesta :
Answer:
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
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
[tex]a= \frac{m2 - m1}{m2+m1}[/tex]
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: brainly.com/question/13376070
