Elevator Physics Problems and Solutions
Some problems on elevators in physics are provided with detailed solutions for high school and college students.
Elevators in Physics:
In all questions involving elevators in physics, there is usually a scale inside the elevator with a person standing on it. They ask us to find the tension force in the cable that supports the elevator's cabin or the scale reading.
Elevators can either accelerate upward or downward; in each case, there are two situations for the elevator's movement.
The elevator can either move upward with increasing speed or downward with decreasing speed. In both cases, the elevator's acceleration is directed upward.
In another case, the elevators can either move upward with decreasing speed or downward with increasing speed. In either case, the acceleration points downward.
In this tutorial, we will explore these situations with solved problems.
Elevators Problems:
Problem (1): An student weighing 550 N stands on a spring scale in a 950-kg elevator (including the person). Just as the elevator starts moving, the scale reads the student's weight as 517 N.
(a) What is the magnitude and direction of the elevator's acceleration?
(b) Find the acceleration if the scale reads 572 N?
(c) What about if the scale reads zero?
Solution: In all elevator problems, the person (or object) exerts a downward force on the scale. According to Newton's third law, the scale also exerts a force on the person standing on it. Additionally, the Earth pulls down on the person.
The force exerted by the scale on the person is called the normal force, denoted by $n$, or apparent weight $w_{app}$ of the object.
(a) To establish a reference direction, we choose the positive $y$-direction to be upward. Since the direction of the elevator's acceleration is not clear, we take its direction as a vector, denoted as $\vec{a}_y$, in the equations. The student's mass can be calculated as follows: \[m=\frac wg=\frac{550}{10}=55\,\rm kg\] There are two forces acting on the student standing on the scale in the elevator: a downward weight force $\vec{w}$ and an upward normal force $\vec{n}$. By drawing a free-body diagram and applying Newton's second law, we obtain: \begin{gather*} \vec{F}_{net}=\vec{w}+\vec{n}=m\vec{a}_y \\\\ -w+n=m\vec{a}_y \\\\ \Rightarrow \boxed{n=w+m\vec{a}_y} \end{gather*} The value that the scale reads is equal to the magnitude of the normal force $N=500\,\rm N$. Substituting this into the equation above yields both the magnitude and direction of the elevator's acceleration: \begin{gather*} n=w+\vec{a}_y \\\\ 517=550+55\vec{a}_y \\\\ \Rightarrow \boxed{\vec{a}_y=-0.6\,\rm m/s^2} \end{gather*} The negative sign indicates that the elevator's acceleration is downward.
It is worth noting that the direction of acceleration does not indicate the direction of motion. In this case, the elevator may either move downward with increasing speed or upward with decreasing speed. In both scenarios, the acceleration would be downward.
(b) In a situation where the scale shows a weight greater than the object's true weight, such as in this case where $n=572\,\rm N$, substituting the numerical values into the following elevator formula yields: \begin{gather*} n=w+m\vec{a}_y \\\\ 572=550+55\vec{a}_y \\\\ \Rightarrow \boxed{\vec{a}_y=+0.4\,\rm m/s^2} \end{gather*} The positive sign indicates that the elevator is accelerating upward. Consequently, in this situation, the student feels heavier than normal.
(c) A zero reading on the scale in a moving elevator signifies that there is no contact force between the scale and the object resting on it. This occurs when the supporting cable of the elevator is broken, causing both the elevator and everything inside it to be in free fall.
Problem (2): An elevator supported by a cable is moving upward and slowing to a stop. By drawing a free-body diagram for this situation, identify all forces acting on the elevator.
Solution: The forces acting on the elevator are as follows: the elevator's weight force, $\vec{w}$, which acts downward, and the upward tension force in the cable supporting the elevator, $\vec{T}$.
The elevator is moving upward and slowing down to a stop. Therefore, according to the definition of acceleration as the change in velocity divided by the time interval, the acceleration would be obtained as a negative value.
Hence, a net force is exerted downward on the elevator. As a result, when drawing the tension and weight forces on the free-body diagram, the weight force must be drawn larger than the tension force to produce a downward net force.
Problem (3): An elevator, hanging from a single cable, has started moving from the upper floors and is speeding up as it descends toward the ground floor. Identify all forces acting on the elevator and then draw a free-body diagram for this situation.
Solution: The elevator is descending with increasing velocity, indicating that its acceleration is directed downward. Consequently, there must be a net downward force exerted on the elevator.
As usual, the tension force $\vec{T}$ in the cable acts upward, while the weight force $\vec{w}$ acts downward on the elevator.
The presence of a net downward force on the elevator implies that the weight force must be greater than the tension force in the free-body diagram depicted in the figure below.
Problem (4): A 68-kg man is standing on a bathroom scale while riding in an elevator that is moving at $\rm 4\,m/s$. As the elevator slows to a stop, the scale shows the man's weight as 710 N.
(a) Was the elevator moving up or down before it stopped?
(b) How long does it take the elevator to come to rest?
Solution: Let's consider the positive $y$-direction as upward. The man's actual weight, $w$, is equal to $68 \times 10 = 680\,\rm N$, which is less than his apparent weight, $w_{app}$, shown by the scale (710 N). Therefore, according to the elevator formula $w_{app}=w+m\vec{a}_y$ his acceleration ($\vec{a}_y$) is positive and directed upward.
(a) The elevator is moving with decreasing speed to come to a stop. This can occur when the elevator is slowing down while moving upward to reach the upper floors or downward to reach the lower floors. In the first case, the acceleration becomes negative; however, in our problem, the acceleration is positive and coincides with the second case. Therefore, the elevator is moving downward at a decreasing speed.
(b) First determine the elevator's acceleration using the following formula \begin{gather*} w_{app}=w+m\vec{a}_y \\\\ 710=680+68\vec{a}_y \\\\ \Rightarrow \boxed{\vec{a}_y=+0.44\,\rm m/s^2}\end{gather*} The final speed of the elevator is zero, $v_f=0$ and it initially moves downward with a velocity of $v_i=-4\,\rm m/s$. Given its acceleration, we can use the following kinematics equation to find the required time the elevator needs to reach a stop. \begin{gather*} v_f=v_i+at \\\\ 0=-4+10t \\\\ \Rightarrow \boxed{t=0.4\,\rm s}\end{gather*}
Problem (5): A woman is standing on a spring scale in an elevator accelerating upward at $1.5\,\rm m/s^2$.
(a) What is the woman's mass, assuming the scale shows $850\,\rm N$?
(b) Considering the total mass of the elevator's cabin plus the scale is $990\,\rm kg$, find the tension in the cable attached to the elevator.
Solution: Assuming up as the positive $x$-direction, the elevator has a positive acceleration of $a_y=+1.5\,\rm m/s^2$. In all problems involving elevators, the value shown on the scale inside is called the apparent weight, $w_{app}$, and it is related to the actual weight $w$ of the object standing on it and the elevator's acceleration vector as \[w_{app}=w+m\vec{a}_y\]
(a) Using the definition of weight, $w=mg$, and substituting the given numerical values into the formula above, we can find the woman's mass as follows: \begin{gather*} w_{app}=m(g+\vec{a}_y) \\\\ 850=m(10+1.5) \\\\ \Rightarrow \boxed{m=74\,\rm kg}\end{gather*}
(b) Two forces are acting on the elevator: a downward weight force (including its weight and everything inside) and an upward tension force in the cable. The resultant of these two forces gives an upward net applied force to the elevator, resulting in an upward acceleration. By applying Newton's second law, we have \begin{gather*}\vec{F}_{net}=\vec{T}+\vec{w}=m\vec{a}_y \\\\ T-w=ma_y \\\\ T-(990\times 10)=990\times 1.5 \\\\ \Rightarrow \quad \boxed{T=11385\,\rm N} \end{gather*} Therefore, the tension in the cable supporting the elevator's cabin is equal to $11385\,\rm N$.
Problem (6): A person is riding in an elevator while standing on a scale. In which situation will the person's apparent weight be the greatest?
Solution: To solve this important question that usually appears in AP Physics exams, we need to consider two key points. First, let's take the positive x-direction upward and then write down the formula that relates the value shown by the scale, which is the apparent weight $w_{app}$, to the elevator's acceleration vector and the object's true weight $w=mg$. \[w_{app}=w+m\vec{a}_y\] We are asked to identify situations in which the person's apparent weight is greater than their actual weight, $w_{app}>w$. This occurs when the second term on the right side of the equation above is added to the true weight, resulting in $w_{app}>w$. Therefore, \begin{gather*} w_{app}=w+m\vec{a}_y \\\\ w_{app}=w+m(+a_y) \\\\ \Rightarrow w_{app}>w \end{gather*} This tells us that for the person's apparent weight to be greater than their actual weight, the elevator should move with a positive acceleration, meaning it is directed upward. This positive acceleration for the elevator can be achieved when it moves with increasing speed upward (speeding up) or decreasing speed downward (slowing down).
Problem (7): A physics student stands on a spring scale in an elevator, initially at rest. Just as the elevator starts to move, it shows only 0.45 of his true weight. What is the magnitude and direction of the elevator?
Solution: As usual, take up the positive $x$-direction and write down the following well-known elevator formula: \[w_{app}=w+m\vec{a}_y\] Here, $w_{app}$ represents the normal force exerted on the person standing on the scale, also known as the person's apparent weight. We are told that $w_{app}=0.45w=0.45(mg)$. Substituting this into the formula above and solving for the acceleration vector $\vec{a}_y$ yields: \begin{gather*} w_{app}=w+m\vec{a}_y \\\\ 0.45w=w+m\vec{a}_y \\\\ -0.55(mg)=m\vec{a}_y \\\\ \Rightarrow \boxed{\vec{a}_y=-0.55\,\rm m/s^2} \end{gather*} Acceleration is defined as a vector in physics. The negative sign indicates its direction towards the negative, or downward, $y$-direction. Therefore, if the elevator accelerates at $0.55\,\rm m/s^2$ downward, the person will feel lighter than normal.
Problem (8): Two forces are applied to an elevator with mass m and supported by a cable: the upward tension force in the cable and the downward weight force due to gravity.
(a) When the elevator is moving upward but accelerating downward, which force is greater, $T$ or $w$?
(b) In the case of moving at a constant velocity, which force is greater, $T$ or $w$?
Solution: Assuming the positive $x$-direction to be upward and applying Newton's second law, we can write an equation that relates the tension in the cable to the elevator's weight and its acceleration vector $\vec{a}_y$: \begin{gather*} \vec{F}_{net}=\vec{w}+\vec{T}=m\vec{a}_y \\\\ -w+T=m\vec{a}_y \end{gather*}
(a) Here, the elevator is moving upward, but its acceleration vector points in the negative $y$-direction, meaning $\vec{a}_y=-a_y$. Substituting this into the formula above gives a comparison between $T$ and $w$ as follows: \begin{gather*} T-w=m\vec{a}_y \\\\ T-w=\underbrace{m(-a_y)}_{<0} \\\\ \Rightarrow \boxed{T<w}\end{gather*} Therefore, in this case, the tension in the cable is less than the cabin's weight.
(b) Constant velocity means no acceleration, i.e., $a_y=0$. Plugging this into the formula above gives \begin{gather*} T-w=m\vec{a}_y=0 \\\\ \Rightarrow \boxed{T=w}\end{gather*} In this situation, the tension of the cable is balanced with the cabin's weight.
The following elevator problems are more relevant for the AP Physics exams or courses.
Problem (9): A 75-kg person stands on a bathroom scale in a motionless elevator. As the elevator rises, it reaches its maximum speed of 1.5 m/s in 0.5 s. The elevator travels at this constant speed for 4 seconds, and while it is about to stop, it undergoes a negative uniform acceleration for 1 second. In each of the following, find the reading of the scale:
(a) Before the elevator begins to move.
(b) During the 0.6 seconds that the elevator was going up.
(c) While the elevator was moving at a constant speed.
(d) Once the elevator was experiencing negative acceleration.
Solution: The scale in a moving elevator always reads the object's apparent weight, which is determined by the following equation: \[w_{app}=w+m\vec{a}_y\] where $\vec{a}_y$ indicates the elevator's acceleration vector.
(a) Before the elevator starts moving, it is at rest and has no acceleration ($\vec{a}_y=0$). Thus, the scale reads the true (actual) weight of the person. \begin{align*}w_{app}&=w=mg \\ &=75\times 10 \\&=750\,\rm N\end{align*}
(b) During this 0.6-second time interval, the elevator starts from rest ($v_i=0$) and accelerates upward to reach its maximum speed of $v_f=1.5\,\rm m/s$. Recall from kinematic equation problems that the relevant equation for solving its acceleration is as follows: \begin{gather*} v_f=v_i+at \\ 1.5=0+a(0.5) \\ \Rightarrow \quad a=+0.3\,\rm m/s^2\end{gather*} Assuming upward to be the positive $y$-direction, the negative sign tells us that the elevator accelerates upward, as expected.
Now that the acceleration vector is known, the reading of the scale is determined as below: \begin{align*} w_{app}&=w+m\vec{a}_y \\\\ &=75\times 10 + 75\times (0.3) \\\\ &=\boxed{772.5\,\rm N} \end{align*}
(c) During this phase, the elevator has no acceleration due to its constant speed. Therefore, the scale shows the apparent weight as the same true weight when the elevator is initially at rest.
(d) The elevator is moving upward at a constant speed for $4\,\rm s$ to finally come to a stop. Therefore, during this time interval, we take the elevator's initial speed as $v_i=1.5\,\rm m/s$ and its final velocity as $v_f=0$. By considering these changes in speed, the acceleration is determined as follows: \begin{gather*} v_f=v_i+at \\ 0=1.5+a(4) \\ \Rightarrow a=-0.37\,\rm m/s^2\end{gather*} The negative sign indicates that the elevator's acceleration is downward, $\vec{a}_y=-0.37\,\rm m/s^2$, but it is moving upward.
Before computing the person's apparent weight, we expect the scale to show a value less than the true weight because its motion in this course is slowing down. \begin{align*} w_{app}&=w+m\vec{a}_y \\\\ &=750+75\times (-0.37) \\\\ &=\boxed{722.25\,\rm N} \end{align*}
Author: Dr. Ali Nemati
Date Published: August 5, 2023
