A uniform meter sick of mass 200 g is held horizontally by two vertical strings, one at the zero cm mark and the other at the 89 cm mark. Find the tension in the string at the 89 cm mark.
T * .89 = .2 * g * .5
T = .1 (9.81) /.89
T = 1.1 Newtons
To find the tension in the string at the 89 cm mark, we can make use of the concept of torque. Torque is the rotational equivalent of force and is given by the product of the force and the perpendicular distance from the axis of rotation.
In this case, the axis of rotation is the center of mass of the meter stick, and the two strings are applying forces on the meter stick. Let's assume that the tension in the string at the zero cm mark is T0, and we need to find the tension in the string at the 89 cm mark, which we'll call T89.
First, we need to determine the center of mass of the meter stick. Since it is a uniform meter stick of mass 200 g, the center of mass will be located at the halfway point, which is 50 cm from either end.
Now, let's calculate the torque about the center of mass due to the tension in the string at the zero cm mark. Since the distance from the center of mass to the zero cm mark is 50 cm, the torque is given by:
Torque0 = T0 * 50 cm
Next, we need to calculate the torque about the center of mass due to the tension in the string at the 89 cm mark. The distance from the center of mass to the 89 cm mark is 39 cm (since the center of mass is already 50 cm from the zero cm mark). So, the torque is given by:
Torque89 = T89 * 39 cm
Since the meter stick is in equilibrium (not rotating), the sum of the torques about any point must be zero. Therefore:
Torque0 + Torque89 = 0
Substituting the values, we get:
T0 * 50 cm + T89 * 39 cm = 0
We know that the weight of the meter stick (mg) also acts at the center of mass and contributes to the torque. Since it is a horizontal stick, the weight is balanced by the tension in the two strings. We can neglect the weight for now, as it doesn't contribute to the net torque.
Since the stick is not rotating, the sum of the two tensions is equal to the weight of the meter stick, which is given by:
T0 + T89 = mg
Substituting the values, we have:
T0 + T89 = (200 g) * (9.8 m/s^2)
Now, we can solve these two equations to find the tension in the string at the 89 cm mark (T89).