When two objects of unequal mass are hung vertically over a frictionless pulley of negligible mass, as in Figure 5.14a, the arrangement is called an Atwood machine. The device is sometimes used in the laboratory to measure the free-fall acceleration. Determine the magnitude of the acceleration of

Question

When two objects of unequal mass are hung vertically over a frictionless pulley of negligible mass, as in Figure 5.14a, the arrangement is called an Atwood machine. The device is sometimes used in the laboratory to measure the free-fall acceleration. Determine the magnitude of the acceleration of the two objects and the tension in the lightweight cord.

Accepted Answer

Conceptualize the situation pictured in Figure 5.14a—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.
The objects in the Atwood machine are subject to the gravitational force as well as to the forces exerted by the strings connected to them—thus, we can categorize this as a Newton’s second law problem. To analyze the situation, the free-body diagrams for the two objects are shown in Figure 5.14b. Two forces act on each object: the upward force T exerted by the string and the downward gravitational force. In problems such as this in which the pulley is modeled as massless and frictionless, the tension in the string on both sides of the pulley is the same. If the pulley has mass and/or is subject to friction, the tensions on either side are not the same and the situation requires techniques we will learn in Chapter 10.

We must be very careful with signs in problems such as this. In Figure 5.14a, notice that if object 1 accelerates upward, then object 2 accelerates downward. Thus, for consistency with signs, if we define the upward direction as positive for object 1, we must define the downward direction as positive for object 2. With this sign convention, both objects accelerate in the same direction as defined by the choice of sign. Furthermore, according to this sign convention, the y component of the net force exerted on object 1 is [latex] T-m_{1} g [/latex], and the y component of the net force exerted on object 2 is [latex] m_{2} g-T [/latex]. Notice that we have chosen the signs of the forces to be consistent with the choices of signs for up and down for each object. If we assume that [latex] m_{2}>m_{1} ext {, then } m_{1} ext { must accelerate upward, while } m_{2} [/latex] must accelerate downward.

When Newton’s second law is applied to object 1, we obtain

[latex] ext { (1) } \quad \sum F_{y}=T-m_{1} g=m_{1} a_{y} [/latex].

Similarly, for object 2 we find

[latex] ext { (2) } \quad \sum F_{y}=m_{2} g-T=m_{2} a_{y} [/latex].

When (2) is added to (1), T cancels and we have

[latex] -m_{1} g+m_{2} g=m_{1} a_{y}+m_{2} a_{y} [/latex].

(3) [latex] a_{y}=\left(\frac{m_{2}-m_{1}}{m_{1}+m_{2}} ight) g [/latex].

The acceleration given by (3) can be interpreted as the ratio of the magnitude of the unbalanced force on the system [latex] \left(m_{2}-m_{1} ight) g ext {, to the total mass of the system }\left(m_{1}+m_{2} ight) [/latex], as expected from Newton’s second law.
When (3) is substituted into (1), we obtain

[latex] ext { (4) } \quad T=\left(\frac{2 m_{1} m_{2}}{m_{1}+m_{2}} ight) g [/latex].

Finalize this problem with the following What If?

Question 5.9: When two objects of unequal mass are hung vertically over a ...

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