# Question #5bc76

May 10, 2016

Apparent weight is the sum of all the forces acting on an object, including its own weight. Apparent weight, ${F}_{A} = \Sigma {F}_{i}$, is defined the same way in the cases of lift and buoyant force.

#### Explanation:

The apparent weight, ${F}_{A}$, of an object can be thought of as the net force acting on an object where the object’s weight is included as one of the forces.

In the case of an object experiencing lift, the apparent weight is

${F}_{A} = \Sigma {F}_{i} = {F}_{W} + {F}_{L}$,

where ${F}_{W}$ is the weight of the object and ${F}_{L}$ is the force providing lift. **

In the case of an object floating in a fluid, the apparent weight is

${F}_{A} = \Sigma {F}_{i} = {F}_{W} + {F}_{B}$,

where ${F}_{W}$ is the weight of the object and ${F}_{B}$ is the buoyant force acting on the object.

In the case of a person standing on a moving elevator.

${F}_{A} = \Sigma {F}_{i} = {F}_{W} + m {a}_{\text{E}}$ where ${F}_{W}$ is the weight of the person, m is the mass of the person and ${a}_{\text{E}}$ the acceleration of the elevator. ${a}_{\text{E}}$ is positive when the elevator accelerates upward and negative when the elevator accelerates downward (e.g. under the acceleration of gravity).

So with that said, your next thought might be the case where the apparent weight ${F}_{A} = 0$. What’s that mean and what does it look like?

Well since ${F}_{W} = m g$, when the elevator or let’s say an airplane is accelerating downward at ${a}_{\text{E}} = - g$ (in other words falling in a nose dive), ${F}_{A} = {F}_{W} + m {a}_{\text{E}} = m g - m g = 0$. Under this condition where ${F}_{A} = 0$ you feel weightless and here’s what it looks like!

* Zero-g, Parabolic Flight.* **

Hope that helps. Good luck!

**Each ${F}_{i}$ represents a different force, $i$ allows us to denote and distinguish them. $i$ typically takes on integer values (e.g. $i = 0 , 1 , 2 , 3$) , but in this case there are only two forces ${F}_{W} + {F}_{L}$, so i takes on the values $i = W$ and $i = L$.