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Finding apparent weight using normal force. |
One way that we perceive weight is the normal force we experience from the ground. In physics problems, when you are asked to determine apparent weight, the quickest method is usually to compute the normal force provided by the "ground".
One way to experience a reduced apparent weight is to strap into a harness of ropes and have someone (or some weight) pull down on the other end like they do in theater or films. Another way is to jump into a swimming pool, where the water lifts up on you. Another possibility, which we explore in this problem, is to enter an environment where the "ground" is capable of moving, such as an elevator.
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{excerpt:hidden=true}Finding apparent weight using normal force.{excerpt}
One way that we perceive weight is the normal force we experience from the ground. In physics problems, when you are asked to determine apparent weight, the quickest method is usually to compute the normal force provided by the "ground".
One way to experience a reduced apparent weight is to strap into a harness of ropes and have someone (or some weight) pull down on the other end like they do in theater or films. Another way is to jump into a swimming pool, where the water lifts up on you. Another possibility, which we explore in this problem, is to enter an environment where the "ground" is capable of moving, such as an elevator.
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{card:label=Part A}
h2. Part A
Suppose a person with a weight of 686 N is in an elevator which is accelerating downwards at a rate of 3.0 m/s{color:black}{^}2{^}{color}. What is the person's apparent weight?
h4. Solution
{toggle-cloak:id=Asys} *System:* {cloak:id=Asys}Person as a [point particle].{cloak}
{toggle-cloak:id=Aint} *Interactions:* {cloak:id=Aint}External influences from the earth (gravity) and the floor of the elevator (normal force).{cloak}
{toggle-cloak:id=Amod} *Model:* {cloak:id=Amod}[Point Particle Dynamics].{cloak}
{toggle-cloak:id=Aapp} *Approach:*
{cloak:id=Aapp}
{toggle-cloak:id=AFBD} {color:red}{*}Diagrammatic Representations{*}{color}
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The physical picture and free body diagram for the person is:
| !elevator1.gif! | !elevator2.gif! |
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{cloak:AFBD}
{toggle-cloak:id=Amath} {color:red}{*}Mathematical Representation{*}{color}
{cloak:id=Amath}
which leads to the form of [Newton's 2nd Law|Newton's Second Law] for the _y_ direction:
{latex}\begin{large}\[ \sum F_{y} = N - mg = ma_{y} \]\end{large}{latex}
In our coordinates, the acceleration of the person is _a{_}{~}y~ = \-3.0 m/s{color:black}{^}2{^}{color}, giving:
{latex}\begin{large}\[ N = ma_{y} + mg = \mbox{476 N} \]\end{large}{latex}
{tip}Is it clear why the acceleration must have a minus sign?{tip}
{tip}This result for the normal force is less than the person's usual weight, in agreement with our expectation that the person should feel lighter while accelerating downward.
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h2. Part B
Suppose a person with a weight of 686 N is in an elevator which has been ascending at a constant rate of 1.0 m/s and is now slowing down at a rate of 3.0 m/s{color:black}{^}2{^}{color}. What is the person's apparent weight?
h4. Solution
{toggle-cloak:id=Bsys} *System, Interactions and Model:* {cloak:id=Bsys}As in Part A.{cloak}
{toggle-cloak:id=Bapp} *Approach:*
{cloak:id=Bapp}
As in Part A, the acceleration is negative in our coordinates. (Why?) The free body diagram is also the same, and so we find the same result:
{latex}\begin{large}\[ N = \mbox{476 N} \]\end{large}{latex}
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{card:label=Part C}
h2. Part C
Suppose a person with a weight of 686 N is in an elevator which is ascending, speeding up at a rate of 3.0 m/s{color:black}{^}2{^}{color}. What is the person's apparent weight?
h4. Solution
{toggle-cloak:id=Csys} *System, Interactions and Model:* {cloak:id=Csys}As in Part A.{cloak}
{toggle-cloak:id=Capp} *Approach:*
{cloak:id=Capp}
The free body diagram and form of Newton's 2nd Law is the same as in Part A, except that the relative size of the forces will be different. We can see this by writing Newton's 2nd Law for the y-direction:
{latex}\begin{large}\[ N = ma_{y} + mg \]\end{large}{latex}
This time, however, the acceleration is positive (Why is this?) Notice that the elevator's velocity does not affect the result, and does not appear in the formula. The acceleration is (_a{_}{~}y~ = + 3.0 m/s{color:black}{^}2{^}{color}) giving:
{latex}\begin{large}\[ N = \mbox{896 N} \] \end{large}{latex}
{tip}Upward acceleration increases the perceived weight.
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| Part BSuppose a person with a weight of 686 N is in an elevator which has been ascending at a constant rate of 1.0 m/s and is now slowing down at a rate of 3.0 m/s2. What is the person's apparent weight? Solution System, Interactions and Model: Approach: |
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| Part CSuppose a person with a weight of 686 N is in an elevator which is ascending, speeding up at a rate of 3.0 m/s2. What is the person's apparent weight? Solution System, Interactions and Model: Approach: |
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