Weight vs. Apparent Weight Physics 11

Elevator:  When you enter the elevator and press the button, you feel the normal amount of your weight on your feet.  When the elevator accelerates upward, you feel slightly heavier than normal. As the elevator continues up at constant speed you feel your normal weight again, but as the elevator slows to stop at your floor you feel slightly lighter than normal.  Why do you feel different things as the elevator’s motion changes?

Weight vs. Apparent Weight:  What you feel, is your apparent weight (F N )  It is not the pull of gravity down on you that you feel as your weight but it is the floor pushing up on you more or less that you feel  In other words, the apparent weight you feel is actually the normal force of the elevator  The normal force changes due to the motion of the elevator floor.

 The equation for net force from the free body diagram:  Rearranging the equation, apparent weight (F N ) is given by the equation:

 If you are accelerating up (Fnet > 0) your apparent weight will be greater.  If you are accelerating down (Fnet < 0) your apparent weight will be smaller  When you are at rest or moving at constant speed (Fnet =0) your apparent weight will be your normal weight.

Elevator Problem:  A person with a mass of 85kg steps onto an elevator  What would a Newton scale read?  If the elevator accelerates up at 2.5m/s 2, what is the new scale reading?  If the elevator slows at - 1.5m/s 2, what is the new scale reading?

Not moving:  F g = mg = (85)(-9.81 m/s 2 ) = N  F net = F g + F N 0 = (-830) +F N 830N = F N

Scale Reading when accelerating at 2.5m/s 2 :  F g = mg = (85)(-9.81 m/s 2 ) = N  F net = ma = (85)(2.5 m/s 2 ) = 210 N  F net = F g + F N 210 = (-830) +F N 1040N = F N

Scale Reading when accelerating at 1.5m/s 2 :  F g = mg = (85)(-9.81 m/s 2 ) = N  F net = ma = (85)(-1.5 m/s 2 ) = -128 N  F net = F g + F N -128 = (-830) +F N 702N = F N

Try it:  A dog with a mass of 34.5 kg steps onto an elevator  What would a Newton scale read?  If the elevator accelerates up at 0.555m/s 2, what is the new scale reading?  If the elevator slows at m/s 2, what is the new scale reading?  Page 186 # 21-23

Free Fall

 While in free fall, all objects are experiencing the force of gravity (assuming zero air resistance) and therefore all accelerate at the same rate  Therefore, objects appear “weightless” because there is no normal force to allow read from a scale or to perceive as weight