Weight, Mass, and the

Dreaded Elevator Problem

Mini-lab: Weight vs. Mass

Determine the mathematical relationship between an object’s weight and its mass.– Materials:

Electronic Scales Triple Beam Balance Multiple objects of different mass

Create a data table BEFORE starting the lab

Follow the instructions at the lab station.

Create a graph of force vs. mass Compare the slope of your graph to g

Weight: True or False?In your journal, re-write the statements below, state whether you think it is true or false, and provide 1-2 sentences of your reasoning.

–The mass of an object depends on its location

–The weight of an object depends on its location

–Mass and weight are the same, but with different units.

Weight vs. Mass

Weight: The force that gravity exerts on an object with mass (m). – This force is what causes falling bodies

to accelerate at 9.80 m/s2.– Weight is ALWAYS directed toward the

center of the earth (down)

– Remember, g = 9.80 m/s2

– Units = Newtons

gmFg

Apparent Weight

Apparent weight is the weight something appears to have as a result of an acceleration.

For example, if you were standing on a scale in an elevator, your apparent weight is the weight the scale would read.

So now for some conceptual practice…

Apparent Weight Practice…

Suppose you have a jet-powered flying platform that can move straight up and down. For each of the following cases, is you apparent weight equal to, greater than, or less then your true weight? Explain.– You are ascending and speeding up– You are descending and speeding up– You are ascending at a constant speed– You are ascending and slowing down– You are descending and slowing down– You are descending at a constant speed

HeavierLighterSame

Constant Vertical Velocity

Example: A leaf falling at terminal velocity Up is the positive (+) direction

+ (use the definition of weight)

F up = Fair

F g

Calculating Apparent Weight

Apparent weight can easily be calculated using the concept of net force.

For example, if you are standing on a scale when you are at rest, what forces are acting on you?– The force of gravity (your weight) and the force

of the scale pushing back up (the normal force) What is the net force in this situation?

– 0 N … you’re in static equilibrium

Draw a Free-body diagram for this situation:

Write out the vector equation:

Fscale = apparent weight

Fg = m g

gscalenet FFF

gscalenet FFF Since this situation is in equilibrium,

Therefore,which means the scale is reading the “True weight”

If the person standing on the scale has a mass of 65.0 kg, what is his weight?

gscale

net

FF

F

0

N637)s

m80.9)(kg0.65( 2 gmFg

Accelerating UpwardsExample: A crate being lifted by a rope

Up is the positive (+) direction

+

(using 2nd Law and the definition of weight)

F up = FT

F g

An Accelerating elevator…

If the elevator is accelerating upwards or downwards, then our problem becomes slightly longer…

For example, let’s say the elevator is accelerating upwards at a rate of 2.00 m/s2. What is now different from our first example?

Draw a free body diagram, including a vector off to the side indicating the direction of the net force:

Then write the vector equation:

Fscale = apparent weight

Fg = m g

Fnet

gscalenet FFF

Since this situation is NOT in equilibrium, the following is ALSO TRUE:

Using substitution, we can determine the size of the apparent weight (the reading on the scale):

gscalenet FFF

maFnet

N767)s

m80.9s

m00.2)(kg0.65(

)(

22

scale

scale

scale

F

gammgmaF

mgFma

Accelerating DownwardsExample: A sky diver in free fall

Up is the positive (+) direction

+

(using 2nd Law and the definition of weight)

F up = Fair

F g

Another Accelerating elevator…

Now let’s say the elevator is accelerating downwards at a rate of 2.00 m/s2. Draw the free-body diagram for this situation:Fscale = apparent

weight

Fg = m gFnet

Again, we can write the vector equations…HOWEVER: the net force is now DOWN, so it (and the acceleration) is therefore a negative value…

Using substitution, we can determine the size of the apparent weight (the reading on the scale):

N507)s

m00.2s

m80.9)(kg0.65(

)()()(

)(

22

scale

scale

scale

F

agmgammgamF

mgFam

)( amF

mgFF

net

scalenet

Now You Try! Determine the apparent weight of a 67 kg

man standing in an elevator when the elevator is:– At rest

– Ascending and speeding up at a rate of 1.5 m/s2

– Ascending and slowing down at a rate of -1.2 m/s2

N660)s

m80.9)(kg67( 2 gmFg

N760)s

m80.9s

m5.1)(kg67(

)(

22

scale

scale

F

gammgmaF

N580)s

m2.1s

m80.9)(kg67(

)()()(

22

scale

scale

F

agmgammgamF