Question 5.12: Two objects with the same magnitude of acceleration Figure 5......

IDENTIFY and SET UP The glider and weight are accelerating, so again we must use Newton’s second law. Our three target variables are the tension T in the string and the accelerations of the two objects.

The two objects move in different directions-one horizontal, one vertical – so we can’t consider them to be a single unit as we did the objects in Example 5.11. Figures 5.15b and 5.15c show our free-body diagrams and coordinate systems. It’s convenient to have both objects accelerate in the positive axis directions, so we chose the positive y direction for the lab weight to be downward.

We consider the string to be massless and to slide over the pulley without friction, so the tension T in the string is the same throughout and it applies a force of the same magnitude T to each object. (You may want to review Conceptual Example 4.10, in which we discussed the tension force exerted by a massless rope.) The weights are m_{1} g and m_{2} g.

While the directions of the two accelerations are different, their magnitudes are the same. (That’s because the string doesn’t stretch, so the two objects must move equal distances in equal times and their speeds at any instant must be equal. When the speeds change, they change at the same rate, so the accelerations of the two objects must have the same magnitude a.) We can express this relationship as a_{1 x}=a_{2 y}=a, which means that we have only two target variables: a and the tension T.

What results do we expect? If m_{1}=0 (or, approximately, for m_{1} much less than m_{2} ) the lab weight will fall freely with acceleration g, and the tension in the string will be zero. For m_{2}=0 (or, approximately, for m_{2} much less than m_{1} ) we expect zero acceleration and zero tension.

EXECUTE Newton’s second law gives

\begin{aligned} \text { Glider: } & \sum F_{x}=T=m_{1} a_{1 x}=m_{1} a \\ \text { Glider: } & \sum F_{y}=n+\left(-m_{1} g\right)=m_{1} a_{1 y}=0 \\ \text { Lab weight: } & \sum F_{y}=m_{2} g+(-T)=m_{2} a_{2 y}=m_{2} a \end{aligned}

(There are no forces on the lab weight in the x-direction.) In these equations we’ve used a_{1 y}=0 (the glider doesn’t accelerate vertically) and a_{1 x}=a_{2 y}=a.

The x-equation for the glider and the equation for the lab weight give us two simultaneous equations for T and a :

\begin{aligned} \text { Glider: } & T=m_{1} a \\ \text { Lab weight: } & m_{2} g-T=m_{2} a \end{aligned}

We add the two equations to eliminate T, giving

m_{2} g=m_{1} a+m_{2} a=\left(m_{1}+m_{2}\right) a

and so the magnitude of each object’s acceleration is

a=\frac{m_{2}}{m_{1}+m_{2}} g

Substituting this back into the glider equation T=m_{1} a, we get

T=\frac{m_{1} m_{2}}{m_{1}+m_{2}} g

EVALUATE The acceleration is in general less than g, as you might expect; the string tension keeps the lab weight from falling freely. The tension T is not equal to the weight m_{2} g of the lab weight, but is less by a factor of m_{1} /\left(m_{1}+m_{2}\right). If T were equal to m_{2} g, then the lab weight would be in equilibrium, and it isn’t.

As predicted, the acceleration is equal to g for m_{1}=0 and equal to zero for m_{2}=0, and T=0 for either m_{1}=0 or m_{2}=0.

CAUTION Tension and weight may not be equal It’s a common mistake to assume that if an object is attached to a vertical string, the string tension must be equal to the object’s weight. That was the case in Example 5.5, where the acceleration was zero, but it’s not the case in this example! (Tension and weight were also not equal for the accelerating lift in Example 5.8.) The only safe approach is always to treat the tension as a variable, as we did here. \|

KEYCONCEPT If two objects are connected by a string under tension, both objects have the same magnitude of acceleration but may accelerate in different directions. Choose the positive x-direction for each object to be in the direction of its acceleration.