Ou are designing a high speed elevator for a new skyscraper...

ou are designing a high-speed elevator for a new skyscraper. The elevator will have a mass limit of 2400 kg (including passengers). For passenger comfort, you choose the maximum ascent speed to be 18.0 m/s, the maximum descent speed to be 10.0 m/s, and the maximum acceleration magnitude to be 5.10 m/s2. Ignore friction. What is the minimum time it will take the elevator to ascend from the lobby to the observation deck, a vertical displacement of 640 m? What is the maximum value of a 60.0-kg passenger’s apparent weight during the ascent? What is the minimum value of a 60.0-kg passenger’s apparent weight during the ascent? What is the minimum time it will take the elevator to descend to the lobby from the observation deck, a vertical displacement of 640 m?

Transcript

00:01 For part a of this problem, we need to use newton's second law to find the answer.

00:06 But first, let's draw a free body diagram to visualize the scenario.

00:10 Here we have the elevator, and we have the tension force going up, and we have gravity going down.

00:21 Now let's think about the scenario where we would have the maximum amount of force.

00:27 Well, if we want the maximum, then tension force has to be greater than the force of gravity.

00:33 And we know that it can accelerate upwards.

00:37 So we can say that the maximum force of tension will occur when acceleration is positive.

00:54 If we define that positive is up and negative is down.

01:04 So what we can write is the mass of the elevator and the people times the acceleration is equal to the force of tension minus the force of gravity which is mass times gravity and now we get that mass times acceleration plus gravity gives us tension force over here all we did was factor out the mass since we have it in both terms now we can simply substitute the information we know that this is how much we have for the elevator and the people inside of it we know that we are actually reading upwards with 1 .2 meters per second square.

02:04 And then we have gravity.

02:07 And when we put this in a calculator, we get this much force.

02:19 Great.

02:19 Now we have the maximum force.

02:23 So this is a maximum.

02:27 Now let's think about the minimum.

02:30 You might be tempted to say, well, when the elevator is not moving, then the force of tension must match the force of...

02:37 Gravity, so that is the minimum.

02:40 But actually, if the elevator is accelerating downwards, then the force of tension will be less than the force of gravity.

02:48 So now we have that.

02:52 The minimum happens when the force of tension, the minimum in the force of tension happens when the acceleration is negative.

03:06 So we'll have a negative mass times acceleration is equal, to the force of tension minus the force of gravity.

03:19 So we get that the force of tension is equal to the mass times gravity minus the acceleration.

03:33 And again we have all the information we need.

03:37 So we simply substitute and we get that the minimum force is roughly 6 ,000 newtons less.

03:55 So it will be 20 ,6004 newtons.

04:00 And this is our minimum.

04:07 Great, now let's move on to part b.

04:09 Here we can use our kinematics equations.

04:12 And notice that we're being asked for the minimum time.

04:18 So that means that we want to make the trip as fast as possible.

04:24 Now, we also have to remember that the problem told us that we cannot go up faster than 18 meters per second.

04:33 So that means that we'll have to split the problem.

04:35 Motion in 2.

04:37 First we'll have acceleration going up of 1 .2 meters per second square.

04:52 And then we'll have constant velocity of 18 meters per second.

05:03 This first part happens in a time t1 and the second one happens in a time t2.

05:12 So let's set up an equation to represent the entire motion...

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