Question 5.9: The Atwood Machine When two objects of unequal mass are hung...
The Atwood Machine
When two objects of unequal mass are hung vertically over a frictionless pulley of negligible mass as in Figure 5.15a, the arrangement is called an Atwood machine. The device is sometimes used in the laboratory to determine the value of g by measuring the acceleration of the objects. Determine the magnitude of the acceleration of the two objects and the tension in the lightweight string.
Conceptualize Imagine the situation pictured in Figure 5.15a in action: as one object moves upward, the other object moves downward. Because the objects are connected by an inextensible string, the distance one object travels in a given time interval must be the same as the distance the other one travels, and their velocities and 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 strings 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 Figure 5.15b. Two forces act on each object: the upward force \overrightarrow{T} exerted by the string and the downward gravitational force. In problems such as this one 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 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 one. In Figure 5.15a, notice that if object 1 accelerates upward, object 2 accelerates downward. Therefore, 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 T – m_1g, and the y component of the net force exerted on object 2 is m_2g – T.
From the particle under a net force model, apply Newton’s second law to object 1:
(1) \Sigma F_y=T-m_1 g=m_1 a_y
Apply Newton’s second law to object 2:
(2) \Sigma F_y=m_2 g-T=m_2 a_y
Add Equation (2) to Equation (1), noticing that T cancels:
– m_1 g + m_2 g = m_1 a_y + m_2 a_ySolve for the acceleration:
(3) a_y=\left(\frac{m_2-m_1}{m_1+m_2}\right) g
Substitute Equation (3) into Equation (1) to find T:
(4) T=m_1\left(g+a_y\right)=\left(\frac{2 m_1 m_2}{m_1+m_2}\right) g
Finalize The acceleration given by Equation (3) can be interpreted as the ratio of the magnitude of the unbalanced force on the system \left(m_2-m_1\right) g to the total mass of the system \left(m_1+m_2\right), as expected from Newton’s second law. Notice that the sign of the acceleration depends on the relative masses of the two objects; if m_2>m_1, the acceleration is positive, corresponding to downward motion for m_2 and upward for m_1. However, if m_1>m_2, Equation (3) gives a negative acceleration, indicating that m_1 moves downward and m_2 moves upward.