Question 2.7: Elevator ride You stand on a bathroom scale in an elevator a......
Elevator ride
You stand on a bathroom scale in an elevator as it makes a trip from the first floor to the tenth floor of a hotel. Your mass is 50 kg. When you stand on the scale in the stationary elevator, it reads 490 N (110 lb). What will the scale read (a) early in the trip while the elevator’s upward acceleration is 1.0 m/s², (b) while the elevator moves up at a constant speed of 4.0 m/s, and (c) when the elevator slows to a stop with a downward acceleration of 1.0 m/s² magnitude?
Sketch and translate We sketch the situation as shown at right, choosing you as the system object. The coordinate axis points upward with its origin at the first floor of the elevator shaft. Your mass is m_{\mathrm{Y}}=50 \mathrm{~kg} , the magnitude of the force that Earth exerts on you is FE on Y = mYg = 490 N, and your acceleration is F_{\mathrm{E} \text { on } \mathrm{Y}}=m_Y g=490 \mathrm{~N} (the upward velocity is increasing); (b) a_y=0(v is a constant 4.0 m>s upward); and (c) a_y=+1.0 \mathrm{~m} / \mathrm{s}^2 (the upward velocity is decreasing, so the acceleration points in the opposite, negative direction).
Simplify and diagram We model you as a point-like object and represent you as a dot in both the motion and force diagrams, shown for each part of the trip in Figures a, b, and c. On the diagrams, E represents Earth, Y is you, and S is the scale. The magnitude of the downward force that Earth exerts does not change (it equals m_{\mathrm{Y}} \vec{\text{g}} \text {, and neither } m_{\mathrm{Y}} \text { nor } \vec{\text{g}} \text { change). } diagrams and motion diagrams are consistent with each other for each part of the trip. The length of the normal force arrows representing the force that the scale exerts on you changes from one case to the next so that the sum of the forces point in the same direction as your velocity change arrow.
Represent mathematically The motion and the forces are entirely along the vertical y-axis. Thus, we use the vertical y -component form of Newton’s second law [Eq. (2.6)] to analyze the process. There are two forces exerted on you (the system object) so there will be two vertical y-component forces on the right side of the equation: the y component of the force that F_{\text {Eon } \mathrm{Y} y}=-m_{\mathrm{Y}} g Earth exerts on you, and the y-component of the normal force that the scale exerts on you, N_{\text {Son } \mathrm{Y} y}=+N_{\text {Son } \mathrm{Y}}:
a_{\mathrm{Y} y}=\frac{\Sigma F_y}{m_{\mathrm{Y}}}=\frac{F_{\mathrm{E} \text { on } \mathrm{Y} y}+N_{\mathrm{S} \text { on } \mathrm{Y} y}}{m_{\mathrm{Y}}}=\frac{-m_{\mathrm{Y}} g+N_{\mathrm{S} \text { on } \mathrm{Y}}}{m_{\mathrm{Y}}} .
Multiplying both sides by m_{\mathrm{Y}} \text {, we get } a_{\mathrm{Y} y}m_{\mathrm{Y}}= -m_{\mathrm{Y}} g+N_{\text {Son } \mathrm{Y}} \text { We can now move } m_{\mathrm{Y}} g \text { to the left } \text { side: } m_{\mathrm{Y}} a_{\mathrm{Y} y}+m_{\mathrm{Y}} g=N_{\mathrm{S} \text { on } \mathrm{Y}} \text {, or }
N_{\text {Son } Y}=m_{\mathrm{Y}} a_{\mathrm{Y} y}+m_{\mathrm{Y}} g=m_{\mathrm{Y}} a_{\mathrm{Y} y}+490 \mathrm{~N}
\text { Remember that } m_{\mathrm{Y}} g=490 \mathrm{~N} \text { is the magnitude of the }\text { force that Earth exerts on you. The expression for } N_{S \text { on } Y} gives the magnitude of the force that the scale exerts on you.
Solve and evaluate We can now use the last equation to predict the scale reading for the three parts of the trip.
(a) Early in the trip, the elevator is speeding up and its acceleration is a_{\mathrm{Y} y}=+1.0 \mathrm{~m} / \mathrm{s}^2 \text {. During that } time interval, the force exerted by the scale on you should be
N_{\text {Son } Y}=m_Y a_{Y y}+490 \mathrm{~N}= (50 kg)(+1.0 m/s²) + 490 N = 540 N
(b) In the middle of the trip, when the elevator moves at constant velocity, your acceleration is zero and the scale should read:
N_{\mathrm{S} \text { on } \mathrm{Y}}=m_{\mathrm{Y}} a_{\mathrm{Y} y}+490 \mathrm{~N}
=(50 \mathrm{~kg})\left(0 \mathrm{~m} / \mathrm{s}^2\right)+490 \mathrm{~N}=490 \mathrm{~N}
(c) When the elevator is slowing down near the end of the trip, its acceleration points downward and is a_y = -1.0 m/s². Then the force exerted by the scale on you should be
\begin{aligned}N_{\text {Son } Y} & =m_{\mathrm{Y}} a_{\mathrm{Y} y}+490 \mathrm{~N} \\& =(50 \mathrm{~kg})\left(-1.0 \mathrm{~m} / \mathrm{s}^2\right)+490 \mathrm{~N}=440 \mathrm{~N}\end{aligned}When the elevator is at rest or moving at constant speed, the scale reading equals the magnitude of the force that Earth exerts on you. When the elevator accelerates upward, the scale reads more. When it accelerates downward, even if you are moving upward, the scale reads less. What is also important is that the motion and force in Figures a, b, and c are consistent witheach other and the force diagrams are consistent with the predicted scale readings—an important consistency check of the motion diagrams, force diagrams, and math.
Try it yourself: What will the scale read when the elevator starts from rest on the tenth floor and moves downward with increasing speed and a downward acceleration of -1.0 m/s², then moves down at constant velocity, and finally slows its downward trip with an acceleration of +1.0 m/s² until it stops at the first floor?
Answer: 440 N, 490 N, and 540 N.