Practice: The system shown below is pulled up with a constant 100 N. Calculate the tension between the blocks. Calculate the acceleration of the system (using +/- to indicate if it is up or down).

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Intro to Forces + Newton's Laws | 16 mins | 0 completed | Learn |

Force Problems with Motion | 28 mins | 0 completed | Learn |

Forces with Multiple Objects | 26 mins | 0 completed | Learn |

Vertical Forces & Equilibrium | 34 mins | 0 completed | Learn |

More 1D Equilibrium | 16 mins | 0 completed | Learn |

Vertical Forces & Acceleration | 26 mins | 0 completed | Learn |

Landing & Jumping Problems | 19 mins | 0 completed | Learn |

Forces in 2D | 35 mins | 0 completed | Learn |

2D Equilibrium | 25 mins | 0 completed | Learn |

Concept #1: Forces with Vertical Acceleration

**Transcript**

Hey guys in this video I want to go over some basic problems. We're going to have a force in the Y-axis that's going to cause an acceleration in the y axis or the vertical axis. Let's check this out. So first of all remember so force problems any force problems we're going to draw a free body diagram. And. If you have their space you can write FBD the that's the first step and we're going to write that to some of our forces equal ma. And we're going to solve the problem. If you have multiple objects we just do this more times one for each object in all these examples I'm going to use g to be 10 now g is not 10 right. So don't get confused and think you can do that the test. I'm just doing this to make this a little faster. Sometimes professors will actually do that but unless he does it is actually 9.8. At least two sort of four part examples that we're going explore here. So the first one I'd give you four different tensions and I want to know for each one of these tensions What is the acceleration that the block would have. So it's a four kilogram object or block that's in the air and it's pulled vertically up so we're going to call this our attention now in all of these There's also obviously going to be an mg pulling down. And gravity we're using 10 so this is. 40. Going down. This is 60. I don't know what is the acceleration of those block. I'm going to use F=ma why because of force problem and to find acceleration in the force problem or use because of me some of our forces because equals ma. There are two forces someone do this. The acceleration, the mass is four and the acceleration is what I'm looking for. So I'm gonna this is a positive 60 because it's 60 going up ,negative 40. So again a positive 20 equals four a. So a equals positive five meters per second. The fact that this is a positive tells me that this acceleration is going up which should make sense because this force is stronger than this force. So if you think of this as a tug of war the force pulling up wins, this part here is very similar but the force is 20 instead. I want you to pause the video and give this a shot. Very small set up. I'm going to keep going. But hopefully you pause that and you try it yourself to make sure you're following.

So this is a four. I got the same mg of a 40. But this tension now is 20 and a quick glance you can see that this force is actually the strongest force. So you should expect that your acceleration is going to be going down which would be a negative ok, so I'm gonna say that sum all forces equals ma. I have a 20 up 40 down the mass of four. And if you solve for this you get negative 20 equals 4a and a equals negative five. The fact that it's a negative tells me that it's going down. OK. So you can just draw it here as a equals 5 with an arrow down so same a equals negative five. And that's the end of that one. I will do the next one here tension is 40, 40. So I have an mg of 40. And the tension is 40. If you look at these they have the same magnitude these forces cancel. This is called equilibrium and this means the acceleration is going to be simply zero. If you don't quite see it you could actually just do the same steps up here and you're going to get this answer, I have a 40 up and I have a 40 down mass is 4 and I'm looking for the acceleration. This just becomes zero equals 4a so a is just. Zero. OK. I want you to do this one where the tension is zero. So there is no tension. Pause the video do it should be real quick. I'm going to jump right into it here. So 4 I have an mg equals 40. And I don't have tension. So it doesn't actually exist. So when I do the sum of all forces equals ma. All I have is a negative 40 going down. Mass is 4 acceleration is negative 10. This should make sense because if the only force is your mg I get that negative mg equals ma and acceleration is negative g. That's why I got a 10 because we're using 10 for gravity instead of a nine point eight. Otherwise it would have been a nine point eight. OK. So very simple very straightforward I think it's going to get a little more complicated but this is the beginning of it. OK. So a five kilogram block is in the air pulled vertically by a string same set up. Another block is five so it's getting pulled down by an mg of 50 and I'm going to have now give you the acceleration and ask for tension. So just the opposite. What is the tension if the acceleration is accelerating down. So the acceleration would be a negative 3 when you plug it into the equation. So this is I'm looking for a force on a force problem so I'm going to do F equals ma, the forces are positive tension plus negative mg or negative 50, mass is five and accelerations we're looking for. So actually acceleration I have. So it is a negative three because I said it's down with three ok so, I have T equals this is negative 15. This 50 goes over to the other side as a positive. And I get a thirty five Newtons.

And we can kind of look here and see if it makes sense this force is smaller than this force. So if you think of this a tug of war, mg wins because it's pulling harder down. Excuse me. So let's look at this one here. I have a five it's being pulled down with an mg of 50 and now it's accelerating up with the constant four and we would know what is its tension. Again I want to pause the video and try this out. Should be very quick and see if you are following if you're getting this correctly. So I'm going to keep going here. Sum of our forces he calls me I have tension up I have negative 50 down mass is five and acceleration is four, four positive so I get two equals this is 20 plus 50. This goes over to the other side becomes a positive equals 70. I got a positive 70 going up over here. All right. So. This means I had to be a positive 70 because I'm pulling up. But anyway. This is bigger than this. So it makes sense that I am accelerating up. All right. So let's do two more here. You're accelerating down with a constant 10 meters per second. I'm gonna do this one you try out the next one. So your here is 50 and your acceleration is 10 down and I want to know what is the tension. So sum of all forces equals ma. The forces are tension up 50 down mass is five and the acceleration is 10. Look what I get T minus 50 equals negative 50. So T equals negative 50. This goes over to the other side becomes positive 50 and I get tension is zero. So that's kind of weird. But look my acceleration is 10 down. That's gravity right. If I'm accelerating with gravity again nine point eight in the space we're doing 10 if I'm accelerating with gravity because there is no tension. So that kind of makes sense. My acceleration in y axis is my negative g. So it makes sense that this problem this object is in freefall because the only force acting on it is gravity. Therefore tension is zero. So that makes sense. I want you to try this real quick and notice that it's a constant seven meters per second. OK. Pause the video. Give it a shot. I'm going to keep going here. One thing to remember as soon as you see constant velocity we should be thinking acceleration equals zero. Which for forces this means that the object is in equilibrium. So I have an mg pulling down 50, acceleration is actually zero acceleration means equilibrium it also means that the forces cancel. It also means that forces cancel. They cancel exactly. So that means that this tension has to be a 50 as well so that it cancels the fact that this object is moving up with a 7. Doesn't matter. It doesn't mean that a force pulling up is stronger the forces cancel the acceleration zero. Then it keeps moving with constant velocity. It's moving with a 7 because at some point it had seven. It was given seven meters per second was accelerated to seven. But then the forces balanced. OK. So kind of tricky but the force have to be the same. Sum of the forces because. The acceleration is simply zero. I know this because it's a constant velocity. So. Plus T plus negative mg equals zero. T has to equal mg and it's just 50 Newtons. OK. Because this problem is at equilibrium. Anyway that's it for the sheet hopefully make sense, let me know if you've any questions

Practice: The system shown below is pulled up with a constant 100 N. Calculate the tension between the blocks. Calculate the acceleration of the system (using +/- to indicate if it is up or down).

Practice: You are inside a bucket that is connected to a pulley above you by a vertical rope. You pull yourself up by pulling down on the other end of the rope. If the total mass of you plus bucket is 80 kg, how hard must you pull down on the rope to move up with a constant 2 m/s^{2} ?

Practice: A 70 kg diver steps off a board 9 m above the water and falls vertical to the water, from rest. If his downward motion is stopped 2.0 s after his body fist touches the water, what average upward force did the water exert on him?

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Concept #1: Forces with Vertical Acceleration

Practice

Practice

Practice

A skydiver feels "weightless" during free fall because
A) there is no normal force acting on the skydiver during the free fall.
B) there is no gravitational force acting on the skydiver during free fall.
C) there is no net force acting on the skydiver during free fall.
D) the gravitational force acting on the skydiver is canceled by air resistance.

In an elevator with a downward acceleration, the normal force acting on a box of mass m resting on the elevator floor is
A) equal to mg.
B) less than mg.
C) greater than mg.
D) zero.

What should be the acceleration of an elevator that makes you feel weightless? Assume that the positive y-axis points up.

Consider a man standing on a scale which is placed in an elevator. When the elevator is stationary, the scale reading is his weight W. Find S, the scale reading when the elevator is moving downward with acceleration a = 1/6 g.
1. S = 8/7 W
2. S = 6/5 W
3. S = 5/7 W
4. S = 0 m/s2
5. S = 7/6 W
6. S = 4/3 W
7. S = 7/5 W
8. S = 5/6 W
9. S = 6/7 W
10. S = 2/3 W

A 55-kg person steps on a scale in an elevator. The scale reads 460 N. What is the elevator doing?[A] The elevator is in free fall.[B] None of these.[C] It is accelerating downward at 1.44 m/s2.[D] It is stationary.[E] It is accelerating upward at 0.41 m/s2.

The figure shows the velocity graph of a 75 kg passenger in an elevator.1) What is the passenger's apparent weight at t = 1.0 s?2) What is the passenger's apparent weight at t = 5.0s?3) What is the passenger's apparent weight at t = 9.0s?

Look at this applet. It shows an elevator with a small initial upward velocity being raised by a cable. The tension in the cable is constant. The energy bar graphs are marked in intervals of 600 J. Part A What is the mass m of the elevator? Use g=10m/s2 for the magnitude of the acceleration of gravity. Express your answer in kilograms to two significant figures.

A 6 kg bucket of water is being pulled straight up by a string at a constant speed.(a) What is the tension on the string?(b) At a certain point, the speed of the bucket begins to change. The bucket now has an upward constant acceleration of magnitude 3 m/s2. What is the tension in the rope now?(c) Now assume that the bucket has a downward acceleration, with a constant acceleration of magnitude 3 m/s2. What is the tension in the rope?

A window washer pulls herself upward using the bucket-pulley apparatus shown in the figure.The mass of the person plus the bucket is 61 kg. (a) How hard must she pull downward to raise herself slowly at constant speed? (b) If she increases this force by 19%, what will her acceleration be?

The upward normal force exerted by the floor is 620 N on an elevator passenger who weighs 650 N. What is the magnitude of the acceleration?

Two crates, A and B are in an elevator as shown. The mass of crate A is greater than the mass of crate B. As the elevator approaches its destination, its speed decreases (it continues to move downward).In the spaces provided below draw arrows to indicate the direction of the net force on each crate. If the net force on either crate is zero, state so and explain.

Two crates, A and B, are in an elevator as shown. The mass of crate A is greater than the mass of crate B. As the elevator approaches its destination, its speed decreases (it continues to move downward).How does the acceleration of crate A compare to that of crate B? Explain

An ascending elevator, hanging from a cable, is coming to a stop.a. Draw a free-body diagram of the object.b. Draw the vectors starting at the black dot. The location and orientation of the vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded.

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