A one-meter rod is suspended at its middle so that it balances. Suppose one-gram on the rod at the following distances from one end.
20cm 27cm 38cm 44cm 58cm 70cm 83cm 94cm
Where must a one gram weight be hung so that the rod will balances at the 50 centimeter mark
Where must two gram weight be hung so that the rod will balance at the 50 centimeter mark
To solve this problem, we need to understand the concept of torque and equilibrium.
Torque is the rotational equivalent of force. It is calculated by multiplying the force applied by the distance from the pivot point (fulcrum). In this case, the fulcrum is at the middle of the rod.
Equilibrium is a state where the system is balanced and not in motion. In rotational equilibrium, the sum of the torques acting on the object must be zero.
Given that the rod balances at its middle, we know that the sum of the torques acting on it from the left side and the right side are equal. Mathematically, we can express this as:
Torque on the left side = Torque on the right side
Now, let's break down the problem:
1. Where must a one gram weight be hung so that the rod will balance at the 50 centimeter mark?
To balance the rod at the 50 centimeter mark, we need to find the point where the torque on the left side is equal to the torque on the right side.
Let's assume we hang the one gram weight at a distance 'x' from the left end of the rod. We can calculate the torque on each side as follows:
Torque on the left side = (20 cm * 1 gram) + (27 cm * 1 gram) + (38 cm * 1 gram) + (44 cm * 1 gram)
Torque on the right side = (58 cm * 1 gram) + (70 cm * 1 gram) + (83 cm * 1 gram) + (94 cm * 1 gram) + (x * 1 gram)
Setting the torques equal, we have:
(20 + 27 + 38 + 44) = (58 + 70 + 83 + 94 + x)
Simplifying the equation, we have:
(129) = (305 + x)
Rearranging the equation, we find:
x = 129 - 305
x = -176
Since the distance cannot be negative, we conclude that the one gram weight cannot be hung on the left side of the 50 centimeter mark in order to balance the rod. Therefore, it is not possible to hang a one gram weight in such a way that the rod balances at the 50 centimeter mark.
2. Where must a two gram weight be hung so that the rod will balance at the 50 centimeter mark?
Similar to the first question, we need to find the point where the torques on both sides are equal.
Let's assume we hang the two gram weight at a distance 'y' from the left end of the rod. The torques can be calculated as follows:
Torque on the left side = (20 cm * 1 gram) + (27 cm * 1 gram) + (38 cm * 1 gram) + (44 cm * 1 gram)
Torque on the right side = (58 cm * 1 gram) + (70 cm * 1 gram) + (83 cm * 1 gram) + (94 cm * 1 gram) + (50 cm * 2 grams) + (y * 2 grams)
Setting the torques equal, we have:
(20 + 27 + 38 + 44) = (58 + 70 + 83 + 94 + 50*2 + y*2)
Simplifying the equation:
(129) = (395 + 2y)
Rearranging the equation:
2y = 129 - 395
2y = -266
y = -133
Again, we obtain a negative distance, which is not possible. Therefore, it is not possible to hang a two gram weight in such a way that the rod balances at the 50 centimeter mark.
In conclusion, for both a one gram weight and a two gram weight, it is not possible to hang them in a position that will balance the rod at the 50 centimeter mark.