A uniform rod 1.2 m long has weight of 1000 dynes and 2000 dynes at its ends. At what point on the rod can it be balanced horizontally if the rod weighs 2500 dynes?
To determine at what point on the rod it can be balanced horizontally, we need to find the center of mass of the rod. The center of mass is the point where the total weight is evenly distributed, and it can be found by balancing the torques (moments) about any point.
To balance torques, we need to consider the weights acting on the rod. The weight of the rod itself is 2500 dynes, which we assume to act at its center. The weight at one end is 1000 dynes, acting at one end of the rod, and the weight at the other end is 2000 dynes, acting at the other end of the rod.
Let's assume that the distance between the center of mass and the 1000-dyne weight is x, and the distance between the center of mass and the 2000-dyne weight is (1.2 - x). In equilibrium, the torques about the center of mass must balance out.
The torque due to the 1000-dyne weight is 1000 dynes * x, and the torque due to the 2000-dyne weight is 2000 dynes * (1.2 - x). The torque due to the weight of the rod itself is 2500 dynes * (1.2 - x/2).
Setting up the equation for torque balance:
1000 dynes * x + 2000 dynes * (1.2 - x) = 2500 dynes * (1.2 - x/2)
Simplifying the equation:
1000x + 2400 - 2000x = 3000 - 1250x
x + 2400 = 3000 - 1250x
2250x = 600
x = 0.267 meters
Therefore, the rod can be balanced horizontally at a point approximately 0.267 meters from the end where it has a weight of 1000 dynes.