https://youtu.be/26XRV4EnLaA
PHYS 1101: Lecture Twelve, Part Four
Okay, our next question. We have a 75 kilogram passenger in an elevator and that person experiences the velocity versus time graph shown. We’re asked what’s the passenger’s apparent weight at t equals 1.5 seconds. And the question just includes some suggestions here to get you on the right track. So here of course is Newton’s second law. These are the mathematics we can use to solve the problem.
First question, what’s the object? Fairly obvious here, we’re focused on the passenger. The passenger I say, because we’re asked about his or her apparent weight and we’re given his or her mass, so that’s our object. What is the acceleration for this object? Well we’re being asked about this property of the passenger and I’m going to describe that for you in a little bit, what the meaning of apparent weight is. We’re being asked that question at this particular instant, at the time, 1.5 seconds. What’s happening at the time 1.5 seconds? Let’s see.
From our graph there is 2 seconds, 4 seconds, so halfway here would be 1, 1.5 seconds would be roughly there. This graph tells me that my velocity at that instant looks to be approximately 6 meters per second. That’s my velocity at this instant.
What’s my acceleration there? Well acceleration tells me how that velocity is changing. On a velocity versus time plot, the acceleration is telling me how the velocity is changing and it’s telling me then the slope, or I need to look at the slope of that line. Acceleration is defined to be v minus v zero divided by t minus t zero, and this is rise over run on a velocity versus time plot. Because the rise on a velocity plot, rise is going to be velocity, that’s Delta v. Run is going to be time, Delta t, and that is the definition of acceleration.
So just quickly, can we from this graph figure out the value of a? Well I need the slope of this line. The slope is constant here. So from t equals 0 to 2 seconds, I have a constant acceleration, constant slope, meaning every second the velocity is increasing by the same amount.
And what is that acceleration? It’s rise, it’s plus 8 over 2 seconds, so my slope and therefore acceleration is 8 meters per second divided by 2 seconds, which is 4 meters per second squared. Every second my velocity is increasing by 4 meters per second, and it’s positive because the rise is positive, the slope is up to the right, so it’s plus.
Okay, that’s getting us to think about the left side of these equations. What about the right side? What forces are involved? I have to draw a free body diagram of this object, of this person. Here’s the dot for them, to represent them and now, what’s touching this person? What forces are on him or her?
Well the person has some mass therefore has a gravitational force. That’s m times g. A person standing in an elevator… let me try to draw a sketch of that so I can picture drawing my surface around this object. Here they would be standing, looks like with one leg up in the air in an elevator. What else is touching them? Or what is touching them? Well you know what, only this surface. The only additional thing I would add is a normal force.
Okay now, what do you think though about the length that I just drew this normal force? Be sure you visually have a consistent picture between the forces that you draw and the acceleration vector. Well the way I drew these forces, it looks like the acceleration is 0. They look to be the same, opposite, so balancing.
But I know I have a positive acceleration that is 4 meters per second squared. My acceleration is not 0, it’s up with this magnitude. So this is not a good picture. A better picture is perhaps something like that. F(n) has to be bigger than w so that when I add these two together I end up with my upward fnet. Okay?
Is there any trig to do? You know what, there isn’t. All my forces are straight up, down, and that’s often the case in these problems so I don’t have any trig. Ready for step five, make a list of the knowns and the thing that you want.
I do know the mass of my passenger, 75 kilograms. So I could calculate m times g, force of gravity. I know that the acceleration is positive 4 meters per second squared. Let me further be clear that that acceleration is an upward acceleration, so I’m going to call it a sub y. That will help me focus on or realize which of these two equations I want to use to solve the problem. It’s going to be this equation on the right, the vertical.
What do I want? I was asked for the apparent weight of this passenger. What is apparent weight? Let me define it for you.
Okay let me start with a discussion about weight. Our true weight is m times g. Let me jot that down. The true weight is this force due to gravity, it has a magnitude of m times g. Our sensation of weight, when introduced this way ties it very strongly to the force due to gravity, but in truth our sensation of how heavy we are, of our physical feeling of weight actually comes from how hard the ground is pushing up on us.
Normally the load that that force that the ground is supporting, how hard it’s pushing up to support us is equal to the force due to gravity because if we’re just standing there, we’re not accelerating, the normal force would be equal to the force of gravity. So our true weight is our sensation of the floor supporting us, of whatever is supporting us for the case where we’re just standing there. We’re in equilibrium.
When it asks for your apparent weight, again we’re asking about our sensation of weight and what sets that, and it’s always going to be the magnitude of the force that’s supporting us. In other words, what surface or what thing is holding us up?
If we’re standing on the elevator floor it’s the floor of the elevator, and so it’s the magnitude of f(n). If we were hanging from a rope inside this elevator then the rope would be holding us up and supporting us, in which case it would be the tension in the rope that we would focus on for the apparent weight. So we have to identify a variable that will show up in our force equation to represent what we want, and the answer for apparent weight is the magnitude of the force supporting us. It’s the magnitude of the normal force.
Here’s my red boxed statement to emphasize that. If ever you’re asked what the apparent weight of the object is, it’s always going to be the magnitude of the force that’s caused by whatever agent is supporting the object. In an elevator it’s the floor. If I were hanging from a rope it would be the tension in that line instead of the normal force. Okay?
So question 13 should be an easy one for you. What variable represents the magnitude that the force exerts on the object? That’s the variable we need to solve for. Question 14, what starting equation are we going to use?
Again, two choices: the horizontal Newton’s second law or the vertical part of Newton’s second law. Pick the one that the variable we need to focus on is going to show up in. Take the ay equation for example. What should we substitute in for the numerator, the Sigma f(y)? That’s the right side. Question 16 asks you now to think about the left side of this equation. What should you substitute in for a(y)? What is the real vertical acceleration for that object? This is what the motion diagram would predict, or shows you.
And then question 17 is following through with the calculation. Let’s really finally answer the question. What is the passenger’s apparent weight? And then check that with your intuition. Let me go back here to our picture of the elevator.
This elevator where we’re looking at a snapshot in time during which here at this point where our velocity is increasing, so we’re speeding up, this would be if you think about a case where if your velocity is becoming more positive you’re moving up, and you’re speeding up so your acceleration is up. So when would that be? We’ve all ridden in elevators. This would be when you’re on the lowest floor, the ground floor and you’re moving up. You’re starting from rest and the elevator is accelerating to speed up to move you up.
At that instant in an elevator, I suspect all of you if you think about it next time, you would appreciate that you feel heavier while that acceleration is going on. Feeling heavier means that your apparent weight is larger than your actual weight. Your sensation of weight is more.
Your apparent weight then, the normal force, the size of the load that the force, the surface is supporting, this normal force is bigger than the weight. It’s bigger because the normal force not only has to counter the force due to gravity, but the floor has to provide enough extra push to speed you up and to get you going in the upward direction, so you feel heavier.