Apparent weight

When you stand on the ground, a normal force balances the gravitational force and prevents you from accelerating through the floor. This normal force is the reason it “feels like something” to have weight, as opposed to being weightless (like in free fall). Weight is always equal to mg\displaystyle m g, but apparent weight depends on the normal force.


What is the apparent weight of a 72 kg person in an elevator that is accelerating upwards at 1.3 m/s2?

There are two forces acting on the person. We will call the second one the elevator force just to be clear. Our net force equation is

Fnet=Felev+Fg=ma\displaystyle \vec{F}_{\text{net}} = \vec{F}_{\text{elev}} + \vec{F}_{\text{g}} = m \vec{a},

and rearranging this gives us

Felev=maFg=mamg=m(ag)\displaystyle \vec{F}_{\text{elev}} = m \vec{a} - \vec{F}_{\text{g}} = m \vec{a} - m \vec{g} = m \left ( \vec{a} - \vec{g} \right ).

Now we can substitute the values, giving us

Felev=(72kg)(1.3m/s2(9.80m/s2))=799.2N\displaystyle \vec{F}_{\text{elev}} = \left ( 72 \, \text{kg} \right ) \left ( 1.3 \, \text{m/s}^{2} - \left ( - 9.80 \, \text{m/s}^{2} \right ) \right ) = 799.2 \, \text{N},

therefore the apparent weight is 8.0 × 102 N.