The artist Anya Calderona constructs the mobile shown in the figure.(Figure 1) In the illustrated configuration, the mobile is perfectly balanced. Assume the strings and crossbars are massless.

If Anya decides to make the star twice as massive, and not change the length of any crossbar or the location of any object, what does she have to do with the mass of the smiley face to keep the mobile in perfect balance? Note that she may have to change masses of other objects to keep the entire structure balanced. Choose one of the option below:
- make it eight times more massive
- make it four times more massive
- make it twice as massive
- nothing
- impossible to tell

Twice as massive.

Make it twice as massive

twice as massive

To answer this question, we need to consider the concept of torque in a balanced mobile. Torque is the measure of an object's tendency to rotate around an axis. In a balanced mobile, the torques acting on each side must cancel each other out.

We can start by understanding the initial configuration of the mobile. Since the mobile is in perfect balance, the torques acting on each side are equal and opposite. Therefore, the sum of torques on one side is equal to zero.

Now, Anya wants to make the star twice as massive without changing the length of any crossbar or the location of any object. To maintain perfect balance, she needs to adjust the mass of the smiley face accordingly.

Let's analyze the torques in the new configuration. Since the star is now twice as massive, its torque will also be twice as large. To balance this increased torque, the torque exerted by the smiley face must also increase.

To achieve this, we can increase the mass of the smiley face. However, the increase in mass should not be linear (i.e., double the mass) because the length of the crossbar and the location of the smiley face remain the same. Instead, the increase in mass should be proportional to the distance between the smiley face and the axis of rotation of the mobile.

In this case, the smiley face is two crossbars away from the axis of rotation. Since the star is twice as massive, the smiley face needs to be four times more massive to maintain perfect balance. Therefore, the correct option is to make the smiley face four times more massive.