A mobile is made of identical objects of mass m suspended so that the lowest row has one object, the row above two objects, and the row above three objects. All the strings are at 45° angles. What is the average tension in the strings connecting the highest row to the ceiling?
C)mg

Question

A mobile is made of identical objects of mass m suspended so that the lowest row has one object, the row above two objects, and the row above three objects. All the strings are at 45° angles. What is the average tension in the strings connecting the highest row to the ceiling? 
C)mg

Quizplus · Accepted Answer

Let the tension in the strings be T. Consider the top-most object. It is suspended by two strings, each at 45° angles. Therefore, the tension in each string is T/√2. Now consider the second top-most row of objects. The tension in the string connected to the top-most object is T/√2 (since it is at 45°). The tension in each of the other two strings is also T/√2 (since each is at 45° and they are symmetric). Therefore, the total upward force on the second top-most row is 2(T/√2) + 2(T/√2) = 4T/√2. Since there are three objects in this row, the average upward force on each object is (4T/√2)/3 = 4T/(3√2). Similarly, considering the bottom two rows, we can find the average upward force on an object in each row to be (2T/√2)/1 = 2T/√2 and (3T/√2)/2 = 3T/(2√2), respectively. 
Now consider the top row of objects again. The total upward force on this row is (2T/√2) + (4T/(3√2)) = (8T/(3√2)). Since there is only one object in this row, the upward force on it must be (8T/(3√2))/1 = 8T/(3√2). Therefore, the tension in the string connected to this object must be equal to its upward force, which is 8T/(3√2). This is the average tension in the strings connecting the highest row to the ceiling. Simplifying, we get 8T/(3√2) = T, so T = 3√2/8. Therefore, the answer is B.

A Mobile Is Made of Identical Objects of Mass M