Video Transcript
A body of mass 75 kilograms was attached to a string fixed to the ceiling of an elevator. Starting from rest, the elevator began accelerating vertically upward at 44 centimeters per square second. During this time, the tension in the string was 𝑇 sub one. After accelerating, the elevator continued moving at the speed it had gained, at which point the tension in the string was 𝑇 sub two. Finally, the elevator decelerated at 24 centimeters per square second, at which point the tension in the string was 𝑇 sub three. Determine the magnitudes of 𝑇 sub one, 𝑇 sub two, and 𝑇 sub three. Consider the acceleration due to gravity to be 9.8 meters per square second.
We’re actually interested in three separate scenarios here. Let’s identify and sketch each of them. The first scenario is when this body, which has a mass of 75 kilograms, is accelerating vertically upward in the lift at a rate of 44 centimeters per square second. We know that the downwards force that this body exerts is called its weight, and that’s mass times gravity. Defining gravity to be 𝑔, we can say then the weight of this body is 75 times 𝑔. We also know that during this period of motion, the tension in the string is 𝑇 sub one. So we add that to our diagram, acting in the direction opposite to its weight. Let’s define the direction the body is moving to be positive. Then, we can add acceleration to our diagram vertically upward at a rate of 44 centimeters per square second.
But we might notice that we’re working in kilograms, and we’ve given the acceleration due to gravity in meters per square second. So we’re going to convert the acceleration into meters per square second by dividing by 100. And so we can say that the body is moving vertically upward at a rate of 0.44 meters per square second.
Let’s now repeat this and draw a diagram for the second part of the motion. This time, we still have the downwards force of the weight of the body, 75𝑔. And the upward force, the tension in the string, is 𝑇 sub two. Now, we’re also told that after accelerating, the elevator moves at the speed it had gained. In other words, it moves at a constant speed. That means the acceleration of the body during this period of motion must be zero meters per square second.
Let’s now deal with the third and final part of the motion of this body. We start in the same way, but this time the upwards force for tension is given by 𝑇 sub three. Now, something interesting happens here with the acceleration. In fact, we’re told that the elevator is decelerating. So we can do one of two things. If we continue to model the direction in which the body is moving upwards to be positive, then we say that since it’s decelerating, the acceleration is negative 24 centimeters per square second. We could of course reverse the direction that we consider to be positive. But for consistency, let’s keep upwards as positive. We’re also going to go ahead and convert this into meters per square second once again by dividing by 100. And that gives us an acceleration of negative 0.24 meters per square second. And that’s really useful.
Now that we’re working in kilograms, meters, and seconds, the SI units, any forces that we calculate will be in newtons. So how do we calculate those forces, specifically 𝑇 sub one, 𝑇 sub two, and 𝑇 sub three? In fact, we’re going to use Newton’s second law of motion. Often stated as 𝐹 equals 𝑚𝑎, it tells us that the force that acts on an object is equal to the mass of the object multiplied by its acceleration. Now, we’ve written here the sum of the vector force is equal to mass times the vector acceleration. But in fact we’re working in one direction, so we don’t need to include the vector symbol. In each case then, we’re going to apply what we know to this formula.
Since we’re still taking the upward direction to be positive, we can say that the sum of the forces during the first part of the body’s motion must be 𝑇 sub one minus 75𝑔. Now, of course, we also know this is equal to the mass of the body multiplied by its acceleration. That’s 75 multiplied by 0.44. And then we can solve this equation to calculate the magnitude of 𝑇 sub one by adding 75𝑔 to both sides. That gives us 𝑇 sub one equals 75𝑔 plus 75 times 0.44, which if we take 𝑔 to be 9.8 is 768 or 768 newtons.
Let’s repeat this process to calculate the magnitude of 𝑇 sub two. Once again, the sum of the forces is 𝑇 sub two minus 75𝑔. But this is equal to mass times acceleration. Now, acceleration is equal to zero. So what we can say is that this body is in a state of equilibrium. And so 𝑇 sub two minus 75𝑔 is equal to 75 times zero, which is equal to zero. This time, we solve for 𝑇 sub two by adding 75𝑔 to both sides. So 𝑇 sub two is 75 times 𝑔, or 75 times 9.8, which is 735 newtons.
We need to apply this process one final time to calculate the value of 𝑇 sub three. We start in a very similar way. This time, the sum of the forces is 𝑇 sub three minus 75𝑔. This is equal to mass times acceleration, which is now 75 multiplied by negative 0.24. To solve for 𝑇 sub three, we once again need to add 75𝑔. So we find 𝑇 sub three is 75 multiplied by negative 0.24 plus 75𝑔, or 75 times 9.8. If we calculate that, we get a value of 717 newtons. And we can probably infer that what we’ve done is likely to be correct.
We might notice that the tension in the string reduces during each stage of its motion. And this makes a lot of sense if we think about how it feels when we’re traveling in a lift. As a lift accelerates upward, we feel like we’re being pushed down towards the floor. For the object on the string, that will be likely to increase the tension. Then, as the lift decelerates, we feel less pressure in the lift. So, in option three, 𝑇 sub three is going to be less than 𝑇 sub two and 𝑇 sub one. So we have 𝑇 sub one is 768 newtons, 𝑇 sub two is 735, and 𝑇 sub three is 717 newtons.
