a uniform meter stick of mass 100g is pivoted about the 30 cm mark. a mass of 200 g is placed at the 10 cm mark. where should a mass of 50 g be placed so that the meter stick balance horizontally?

The mass of the stick where M resides is 30g, centered at distance 15.

The mass of the stick where m resides is 70, centered at distance 35

Adding up all the mass*distance values, they must balance on both sides of the fulcrum.

30(15) + 200(20) = 70(35) + 50(d)
450 + 4000 = 2450 + 50d
50d = 2000
d = 40

so, m sits at 40cm from the 30cm mark, or at the 70cm mark.

To balance the meter stick horizontally, the torques on both sides of the pivot point must be equal.

The torque is calculated by multiplying the weight of an object by its distance from the pivot point. In this case, the distance is measured from the 30 cm mark.

On one side of the pivot (left side), we have:
Torque_left = (Mass_left * g) * Distance_left

On the other side of the pivot (right side), we have:
Torque_right = (Mass_right * g) * Distance_right

Given:
Mass_left = 200 g
Distance_left = 10 cm = 0.1 m

Mass_right = 100 g
Distance_right = ? (to solve for)

Mass_50g = 50 g
Distance_50g = ? (to solve for)

We can set up the equation for balancing the torques:

Torque_left = Torque_right

(Mass_left * g) * Distance_left = (Mass_right * g) * Distance_right

(200g * 9.8 m/s^2) * 0.1 m = (100g * 9.8 m/s^2) * Distance_right

196 N.m = 98 N.m * Distance_right

Distance_right = 196 N.m / 98 N.m

Distance_right = 2 m

Therefore, the mass of 50 g should be placed at the 2 cm mark in order to balance the meter stick horizontally.

To find the position where a mass of 50g should be placed in order for the meter stick to balance horizontally, we can use the principle of moments.

The principle of moments states that for an object to be in rotational equilibrium, the sum of the clockwise moments must be equal to the sum of the counterclockwise moments.

In this case, let's consider the counterclockwise moments as positive and the clockwise moments as negative.

Since the meter stick is balanced at the 30 cm mark, the distance from the pivot (30 cm mark) to the fulcrum is zero. Therefore, we can ignore it in our calculations.

Now, let's calculate the moments caused by each mass:

1. The moment caused by the 100g mass at the 30 cm mark is 0, as its distance from the fulcrum is zero.

2. The moment caused by the 200g mass at the 10 cm mark can be calculated as:
Moment = mass × distance from fulcrum
Moment = 200g × 10cm

3. The moment caused by the 50g mass at an unknown position can be calculated as:
Moment = 50g × distance from fulcrum

Since the meter stick is in rotational equilibrium, the sum of the moments must be zero:

0 + (200g × 10cm) + (50g × distance from fulcrum) = 0

Now, we can solve for the distance from the fulcrum where the 50g mass should be placed:

(200g × 10cm) + (50g × distance from fulcrum) = 0

(200g × 10cm) = -(50g × distance from fulcrum)

2000g cm = -50g × distance from fulcrum

distance from fulcrum = (2000g cm) / -50g

distance from fulcrum = -40 cm

Therefore, a mass of 50g should be placed at the 40 cm mark in order for the meter stick to balance horizontally.