EXPLORE When two objects of unequal masses are suspended on opposite sides of a

Question

EXPLORE When two objects of unequal masses are suspended on opposite sides of a frictionless pulley with negligible mass, the arrangement is called an Atwood machine. Suppose an Atwood machine has a mass of m1 = 16.9 kg and another mass of m2 = 18.4 kg hanging on opposite sides of the pulley. Determine the magnitude of the acceleration of the two objects and the tension in the lightweight cord. CONCEPTUALIZE Imagine the situation pictured in the Active Figure. As one object moves upward, the other object moves downward. Because the objects are connected by an inextensible string, their accelerations must be of equal magnitude. CATEGORIZE The objects in the Atwood machine are subject to the gravitational force as well as to the forces exerted by the string connected to them. Therefore, we can categorize this problem as one involving two particles under a net force. ANALYZE The free-body diagrams for the two objects are shown in the Active Figure. Two forces act on each object: the upward force T exerted by the string and the downward gravitational force. In problems such as this one the pulley is modeled as massless and frictionless, and the tension in the string is the same on both sides of the pulley. If the pulley has mass or is subject to friction, other techniques must be used to take these factors into account. The signs used in problems such as this require care. When object 1 accelerates upward, object 2 accelerates downward. Therefore, m1 going up and m2 going down should be represented equivalently by the same acceleration with the same sign. We can do that by defining our sign convention with up as positive for m1 and down as positive for m2. With this sign convention, the y-component of the net force exerted on object 1 is T - m1g, and the y-component of the net force exerted on object 2 is m2g - T. Applying Newton's second law to m1 gives the result Do the same thing for m2: The acceleration ay is the same for both objects. When Equation (1) is added to Equation (2), T cancels and we obtain Solving Equation (3) for ay and substituting for the given values of m1 = 16.9 kg and m2 = 18.4 kg gives Substitute Equation (4) into Equation (1) to find T.

Explanation / Answer

let tension be T and acceleration be a downwards for m2

so:

m2g - T = m2a

T - m1g = m1a

=> a = (m2 - m1)/(m1 + m2) = .0424929 m/s2

=> T = m1(g+a) = 166.5071 N

Here I used value of g as 9.81 m/s2

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