A 1 kg mass is hung at the 0 mark of a meter stick and a 3 kg mass is hung at the 100 cm mark of the same meter stick. If the mass of the meter stick is 0.2 kg, at which of the following points of the meter stick could you balance the system on one finger?

Select one:
a. 38.9 cm
b. 50 cm
c. 73.8 cm
d. 96.9 cm

66cm

Let fulcrum be at x cm from the 0 mark.

Then 1kg*x cm = 3kg*(100-x) cm

Expand and solve for x.

To find the point on the meter stick where the system can be balanced, we need to consider the moments or torques acting on the system.

The moment or torque is calculated by multiplying the force by the distance from the pivot point. In this case, the pivot point is the point where you will place your finger to balance the system.

The total clockwise moment or torque must be equal to the total counterclockwise moment or torque for the system to be balanced.

Let's denote the position where you could balance the system on the meter stick as x cm.

Considering the clockwise torques:
1 kg mass: 1 kg * g * (x cm)
Meter stick: 0.2 kg * g * (x cm)

Considering the counterclockwise torques:
3 kg mass: 3 kg * g * (100 cm - x cm)

Setting the clockwise and counterclockwise torques equal:

1 kg * g * (x cm) + 0.2 kg * g * (x cm) = 3 kg * g * (100 cm - x cm)

Simplifying:

1 * (x cm) + 0.2 * (x cm) = 3 * (100 cm - x cm)

x + 0.2x = 300 - 3x

1.2x + 3x = 300

4.2x = 300

x ≈ 71.4 cm

Rounded to the nearest tenth, the point where you could balance the system on one finger is approximately 71.4 cm.

Therefore, the correct answer is c. 73.8 cm.

To find the point on the meter stick where you could balance the system on one finger, you need to consider the torques acting on the system. The torque depends on the distance from the pivot point and the weight of the masses.

The torque equation is given by:

Torque = Force × Distance

In this case, the force is the weight of the mass, and we can calculate it using the equation:

Force = Mass × Acceleration due to gravity

Let's find the torques produced by the 1 kg mass and the 3 kg mass separately at different points on the meter stick.

For the 1 kg mass at the 0 cm mark (the pivot point):

Torque1 = Force1 × Distance1
= (1 kg × 9.8 m/s^2) × (0 cm)

Since the distance from the pivot point is 0 cm, the torque produced by the 1 kg mass at the 0 cm mark is zero.

Similarly, for the 3 kg mass at the 100 cm mark:

Torque2 = Force2 × Distance2
= (3 kg × 9.8 m/s^2) × (100 cm)

Converting cm to meters:
Torque2 = (3 kg × 9.8 m/s^2) × (1 m)
= 29.4 Nm

Now, let's calculate the torque produced by the meter stick itself. The meter stick has a mass of 0.2 kg and is considered to act at its center of mass, which is at the 50 cm mark.

Torque3 = Force3 × Distance3
= (0.2 kg × 9.8 m/s^2) × (50 cm)

Converting cm to meters:
Torque3 = (0.2 kg × 9.8 m/s^2) × (0.5 m)
= 0.98 Nm

To balance the system on one finger, the total torque from the 1 kg mass, the 3 kg mass, and the meter stick should be zero.

Total Torque = Torque1 + Torque2 + Torque3

Now, let's calculate the total torque for each given point on the meter stick:

a. At 38.9 cm:
Total Torque = 0 Nm + 29.4 Nm + Force4 × (38.9 cm)

b. At 50 cm:
Total Torque = 0 Nm + 29.4 Nm + 0.98 Nm

c. At 73.8 cm:
Total Torque = 0 Nm + 29.4 Nm + Force4 × (73.8 cm)

d. At 96.9 cm:
Total Torque = 0 Nm + 29.4 Nm + Force4 × (96.9 cm)

As we want the total torque to be zero for balancing, the only option where the system could balance on one finger is at 50 cm.

Therefore, the correct answer is option b. 50 cm.