Question

A uniform meter stick with a mass of 200 g is supported horizontally by two vertical strings, one at the 0-cm mark and the other at the 90-cm mark(Figure 1). Figure1 of 1 Part A What is the tension in the string at 0 cm? Express your answer to two significant figures and include the appropriate units. F0cm F 0 c m = .19N SubmitPrevious AnswersRequest Answer Incorrect; Try Again; One attempt remaining Part B What is the tension in the string at 90 cm?

Answer 1

Answer 2

To determine the tension in the string at the 0 cm mark and the 90 cm mark of the uniform meter stick, we need to consider the equilibrium of forces acting on the meter stick.

Let's denote the tension in the string at the 0 cm mark as T0 and the tension in the string at the 90 cm mark as T90.

Since the meter stick is in equilibrium, the sum of the clockwise torques about any point must be equal to the sum of the counterclockwise torques about the same point.

For Part A: To find the tension in the string at the 0 cm mark (T0), we consider the torques acting about the 90 cm mark.

Since the meter stick is uniform, we can assume its center of mass is located at the 50 cm mark. The weight of the meter stick (200 g) acts downward at its center of mass.

The torque exerted by the weight is given by: Torque = (Distance from the pivot) * (Force)

Since the 90 cm mark is 40 cm away from the center of mass: Torque due to weight = (40 cm) * (200 g * g)

To maintain equilibrium, the tension at the 0 cm mark must exert an equal and opposite torque. Since the distance from the 90 cm mark to the 0 cm mark is 90 cm, the torque exerted by the tension at the 0 cm mark is: Torque due to tension at 0 cm = (90 cm) * T0

Setting the torques equal to each other: (40 cm) * (200 g * g) = (90 cm) * T0

Simplifying the equation and solving for T0: T0 = (40 cm) * (200 g * g) / (90 cm)

Calculating the value: T0 ≈ 88.89 N

Therefore, the tension in the string at the 0 cm mark is approximately 88.89 N.

For Part B: To find the tension in the string at the 90 cm mark (T90), we consider the torques acting about the 0 cm mark.

Similarly, the torque exerted by the weight is given by: Torque = (Distance from the pivot) * (Force)

Since the 0 cm mark is 50 cm away from the center of mass: Torque due to weight = (50 cm) * (200 g * g)

To maintain equilibrium, the tension at the 90 cm mark must exert an equal and opposite torque. Since the distance from the 0 cm mark to the 90 cm mark is 90 cm, the torque exerted by the tension at the 90 cm mark is: Torque due to tension at 90 cm = (90 cm) * T90

Setting the torques equal to each other: (50 cm) * (200 g * g) = (90 cm) * T90

Simplifying the equation and solving for T90: T90 = (50 cm) * (200 g * g) / (90 cm)

Calculating the value: T90 ≈ 111.11 N

Therefore, the tension in the string at the 90 cm mark is approximately 111.11 N.