Question Video: Analysing the Motion of an Elevator Using a Set of Scales Mathematics • Third Year of Secondary School

Video Transcript

A man was standing on a set of scales in an elevator, recording the readings as the elevator moved. His first reading was when the elevator was accelerating upward at a rate of seven 𝑎. He made another reading when the elevator was accelerating downward at a rate of eight 𝑎. Given that the ratio between the two readings was four to one, determine the ratio 𝑎 to 𝑔, where 𝑔 is the acceleration due to gravity.

In order to answer this question, we’re going to formalize our thoughts by drawing a free-body diagram of our scenario. The first scenario is when the elevator is accelerating upward; it’s accelerating at a rate of seven 𝑎 length units per square time units. So, we’re going to model in this case the upward direction to be positive. He then made another reading when the elevator was accelerating downward, this time at a rate of eight 𝑎 length units per square time units.

And there are two ways that we could model this. In the second scenario, we could model the downwards direction to be positive. Or for consistency, we could model the upward direction to be positive and say that the acceleration is negative eight 𝑎. And we know that the man by himself exerts a downward force on those sets of scales. That force is his weight, and it’s equal to his mass times gravity. Let’s call that 𝑚𝑔. There is a reaction force acting in the opposite direction to the downwards force of the weight. This reaction force is what gives us the reading on that set of scales. So, let’s define the first reaction force and hence the first reading to be 𝑅 sub one and the second reaction force to be 𝑅 sub two.

Then, we’re going to model both scenarios by using Newton’s third law of motion. The vector form of this says the net sum of the forces is equal to the mass times the vector acceleration. Since we’re only working in one direction here, we can simplify this to the scalar equivalent force equals mass times acceleration. For the first scenario where the elevator is accelerating upward, the net force is found by adding 𝑅 sub one and negative 𝑚𝑔. We define that 𝑚𝑔, the weight force, to be negative as it’s acting downwards, which is the negative direction. So, the net sum of the forces is 𝑅 sub one minus 𝑚𝑔. And this is equal to the mass of the man times acceleration, so 𝑚 times seven 𝑎. Let’s simplify that to seven 𝑚𝑎.

We repeat this process for the second scenario when the elevator is accelerating downward. Now, since we’ve still defined the upward direction to be positive, the net force is 𝑅 sub two minus 𝑚𝑔. But this time, mass times acceleration is 𝑚 times negative eight 𝑎. And that’s negative eight 𝑚𝑎.

Now, the question wants us to determine the ratio between 𝑎 and 𝑔. So, actually we need to begin by finding an expression for 𝑎. It also tells us that the ratio between the two readings is four to one. In other words, 𝑅 sub one to 𝑅 sub two is equal to four to one. So, how do we know that the ratio is 𝑅 sub one to 𝑅 sub two and not the other way around? Well, we could determine it directly from the order in which the readings are mentioned. The first reading is mentioned first, so this should represent the value four.

Alternatively, though, we can use a little bit of logic to understand why 𝑅 sub one might be greater than 𝑅 sub two. Imagine we’re standing in a lift. As the lift accelerates upward, we feel a downward force. If we were standing on a set of scales and we were looking directly at the scales, we might notice that as the elevator accelerates, the value on those scales increases. Similarly, if the elevator began to decelerate, the value on the scales would decrease. So, we might deduce that the first reading is greater than the second. So, we have 𝑅 sub one to 𝑅 sub two equals four to one. If we divide each value on the left of our ratio by each value on the right, we get 𝑅 sub one divided by 𝑅 sub two is equivalent to four over one, or simply 𝑅 sub one divided by 𝑅 sub two equals four.

Then, multiplying through by 𝑅 sub two and we get an equation 𝑅 sub one equals four 𝑅 sub two. We’re going to substitute this directly into equation one. Then, we will have two equations involving 𝑅 sub two. So, equation one becomes four 𝑅 sub two minus 𝑚𝑔 equals seven 𝑚𝑎. We’re going to find a way to eliminate 𝑅 sub two. And to do so, we’re going to multiply this second equation by four. When we do, we get four 𝑅 sub two minus four 𝑚𝑔 equals negative 32𝑚𝑎. Let’s call this equation three. And then, we see we can eliminate four 𝑅 sub two from both equations by subtracting one from the other.

We’ve cleared some space, and we’re going to subtract equation three from equation one. When we do, the four 𝑅 sub twos disappear. We then have negative 𝑚𝑔 minus negative four 𝑚𝑔 on the left-hand side and on the right seven 𝑚𝑎 minus negative 32𝑚𝑎. This simplifies to say that three 𝑚𝑔 equals 39𝑚𝑎. And we now notice that there is a constant factor of 𝑚 that we can divide through by. And we’re allowed to do this since 𝑚 is the mass of the man; it cannot be equal to zero. Similarly, we can actually divide through by three. So, we have the equation 𝑔 equals 13𝑎.

Remember, we were trying to determine the ratio of 𝑎 to 𝑔. We certainly have a relationship between 𝑎 and 𝑔, but it’s not yet in ratio form. What we could do though is divide both sides of this equation by 𝑎. So, 𝑔 over 𝑎 is equal to 13. At this stage, what we’re looking to do is reverse the process we applied when we were looking at the ratio of the two readings. We can then write 13 as equivalently 13 over one. Then, by reversing what we did right at the start, we can say that the ratio of 𝑎 to 𝑔 must be equal to the ratio of one to 13. And so, we’ve determined the ratio 𝑎 to 𝑔, where 𝑔 is the acceleration due to gravity; it’s one to 13.