A uniform meter stick ( 100 cm long) that weighs 0.50 N has a 2 N weight located at the 30 cm mark (from the left) and a 5 N weight at the 85 cm mark (from the left). Where along the meter stick is the center of gravity of this system?
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To find the center of gravity of the system, we need to determine the point along the meter stick where the total torque (or moment) exerted by the weights on either side of that point is balanced.
The torque exerted by a weight is given by the product of its weight (force) and the perpendicular distance from the point of rotation (in this case, the left end of the meter stick). Mathematically, the torque (T) can be calculated using the formula:
T = F * d
where T represents the torque, F is the force (weight), and d is the perpendicular distance.
In this problem, we are given the weights (forces) and the distances from the left end of the meter stick. We can calculate the torques exerted by each weight and then find the point where the total torque on either side of that point is balanced.
Let's calculate the torques exerted by the 2 N and 5 N weights:
Torque exerted by the 2 N weight:
T1 = F1 * d1 = 2 N * 30 cm = 60 N·cm
Torque exerted by the 5 N weight:
T2 = F2 * d2 = 5 N * 85 cm = 425 N·cm
Now, to find the point where the torques are balanced, we need to find the position along the meter stick where the sum of the torques on either side of that point is equal.
Let's assume the center of gravity is located at a distance x from the left end of the meter stick. The torque on the left side of the center of gravity is equal to the torque on the right side:
T1 = T2
2 N * x = 5 N * (100 cm - x)
Now we can solve this equation for x:
2x = 500 - 5x
7x = 500
x ≈ 71.43 cm
Therefore, the center of gravity of this system is located approximately at the 71.43 cm mark from the left end of the meter stick.