three masses are suspended from a meter stick as shown. How much mass must be suspended on the right side for the system to be in equilibrium?
"as shown" ?
To determine the mass that must be suspended on the right side for the system to be in equilibrium, we need to consider the principles of torques.
1. First, let's assign some variables:
- The mass on the left side of the meter stick: m₁ (in kg)
- The distance of m₁ from the center of the meter stick: d₁ (in meters)
- The mass on the right side of the meter stick: m₂ (unknown)
- The distance of m₂ from the center of the meter stick: d₂ (in meters)
2. The equilibrium condition for an object in rotational equilibrium is that the sum of the torques acting on it must be zero. Torque (τ) is calculated by multiplying the force acting on an object by the distance from the pivot point (r).
3. In this case, the pivot point is the center of the meter stick, so the torque equation becomes:
Στ = 0
(m₁ * g * d₁) + (m₂ * g * d₂) = 0
Note: g is the acceleration due to gravity, which is approximately 9.8 m/s².
4. Rearranging the equation, we can solve for m₂:
m₂ * g * d₂ = - (m₁ * g * d₁)
m₂ = - (m₁ * g * d₁) / (g * d₂)
5. Simplifying further:
m₂ = - (m₁ * d₁) / d₂
6. Now, plug in the known values for the problem to find the mass required on the right side for equilibrium.