Weight, Mass, and the Dreaded Elevator Problem

Weight: True or False? In your journal, re-write the statements below, state whether you think it is true or false, and show 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: True or False? In your journal, re-write the statements below, state whether you think it is true or false, and show your reasoning. The mass of an object depends on its location - False The weight of an object depends on its location- True Mass and weight are the same, but with different units- False

Weight vs. Mass Weight is a vector= The force that the Earth exerts on an object with mass (m). This force is what causes falling bodies to accelerate at 9.81 m/s2. Weight is ALWAYS directed toward the center of the earth (down) Remember, g = 9.81 m/s2 Units = Newtons, measured with spring scale

Mass Is a scalar= the amount of substance present in an object Measured using triple beam balance, electronic scale Unit: Kg ( kilogram)

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 Heavier Lighter Same

Constant Vertical Velocity Example: A leaf falling at terminal velocity Up is the positive (+) direction 𝑎=0 so… 𝐹 𝑁𝐸𝑇 =0 𝐹 𝑁𝐸𝑇 =0= 𝐹 𝑢𝑝 − 𝐹 𝑔 + 𝐹 𝑔 + 𝐹 𝑔 𝐹 𝑢𝑝 = 𝐹 𝑔 (use the definition of weight) 𝐹 𝑢𝑝 =𝑚𝑔 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

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( g=9.80m/s2)?

Accelerating Upwards Example: A crate being lifted by a rope Up is the positive (+) direction 𝑎 is↑so… 𝐹 𝑁𝐸𝑇 is↑ 𝐹 𝑁𝐸𝑇 = 𝐹 𝑢𝑝 − 𝐹 𝑔 + 𝐹 𝑔 + 𝐹 𝑔 𝐹 𝑁𝐸𝑇 + 𝐹 𝑔 = 𝐹 𝑢𝑝 𝑚𝑎+𝑚𝑔= 𝐹 𝑢𝑝 (using 2nd Law and the definition of weight) 𝐹 𝑢𝑝 =𝑚 𝑎+𝑔 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?

Then write the vector equation: 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

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):

Accelerating Downwards Example: A sky diver in free fall Up is the positive (+) direction 𝑎 is↓so… 𝐹 𝑁𝐸𝑇 is↓ − 𝐹 𝑁𝐸𝑇 = 𝐹 𝑢𝑝 − 𝐹 𝑔 + 𝐹 𝑔 + 𝐹 𝑔 −𝐹 𝑁𝐸𝑇 + 𝐹 𝑔 = 𝐹 𝑢𝑝 −𝑚𝑎+𝑚𝑔= 𝐹 𝑢𝑝 (using 2nd Law and the definition of weight) 𝐹 𝑢𝑝 =𝑚 𝑔−𝑎 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. (g=9.80m/s2). Draw the free-body diagram for this situation: Fscale = apparent weight Fg = m g Fnet

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):

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