A 0.112 kg meterstick is supported at its 38.6 cm mark by a string attached to the ceiling. A 0.655 kg mass hangs vertically from the 4.74 cm mark. A mass is attached somewhere on the meterstick to keep it horizontal and in both rotational and translational equilibrium. The force applied by the string attaching the meter stick to the ceiling is 19.9 N.Find the point where the mass attaches to the

stick.

To find the point where the mass attaches to the meterstick, we can use the principle of torques (moments) in rotational equilibrium.

The principle of torques states that for an object to be in rotational equilibrium, the sum of the torques acting on it must be zero.

First, let's consider the forces acting on the meterstick:

1. The gravitational force acting on the meterstick itself can be considered as acting at its center of mass, which is located at the midpoint of the meterstick. This force can be calculated by multiplying the mass of the meterstick (0.112 kg) by the acceleration due to gravity (9.8 m/s^2). Therefore, the magnitude of the gravitational force acting on the meterstick is:

F_grav_meterstick = (0.112 kg) * (9.8 m/s^2)

2. The tension force exerted by the string attached to the ceiling is given as 19.9 N.

Next, let's consider the forces acting on the mass:

1. The gravitational force acting on the mass is downward and can be calculated by multiplying the mass of the hanging mass (0.655 kg) by the acceleration due to gravity (9.8 m/s^2).

Now, let's calculate the torques:

The torque is the product of the force and the distance from the pivot point (the point where the meterstick is supported by the string) to the line of action of the force.

1. The gravitational torque acting on the meterstick is zero since the center of mass of the meterstick is at the pivot point.

2. The tension force exerted by the string will create a clockwise torque. The distance from the pivot point to the tension force is 38.6 cm.

3. The gravitational force acting on the hanging mass will create a counterclockwise torque. The distance from the pivot point to the hanging mass is 4.74 cm.

Since the meterstick is in rotational equilibrium, we can set up the equation:

(Tension * Distance_tension) - (GravitationalForce_hangingmass * Distance_hangingmass) = 0

Substituting the known values into the equation, we get:

(19.9 N) * (0.386 m) - (0.655 kg * 9.8 m/s^2) * (0.0474 m) = 0

Simplifying the equation yields:

7.686 N - 3.207 N = 0

Therefore, the point where the mass attaches to the stick is such that the clockwise torque created by the tension force in the string is equal to the counterclockwise torque created by the gravitational force acting on the hanging mass.

To solve for this point, we rearrange the equation and isolate the distance to find:

Tension * Distance_tension = GravitationalForce_hangingmass * Distance_hangingmass

Distance_tension = (GravitationalForce_hangingmass * Distance_hangingmass) / Tension

Plugging the known values into the equation, we find:

Distance_tension = (0.655 kg * 9.8 m/s^2 * 0.0474 m) / 19.9 N

By calculating the above expression, we can find the point where the mass attaches to the meterstick.