Practice: The system below does not move. Find the minimum value that μ,S (between 8 kg block and table) can have.

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Intro to Friction | 40 mins | 0 completed | Learn |

Kinetic Friction | 39 mins | 0 completed | Learn |

Static Friction | 15 mins | 0 completed | Learn |

Inclined Planes | 34 mins | 0 completed | Learn |

Inclines with Friction | 73 mins | 0 completed | Learn |

Forces in Connected Objects (Systems) | 50 mins | 0 completed | Learn |

Intro to Springs (Hooke's Law) | 21 mins | 0 completed | Learn |

Concept #1: Static Friction & Equilibrium

**Transcript**

Hey guys, so in this video I want to show a few examples where static friction will be used to keep a system for moving, either a block or a group of blocks, let me show you. So it says here, friction can be used to keep objects from moving or accelerating. In these cases, friction would be cancelling out another force, okay? So, let me show you. Here I have a 10 kilogram block that is being pushed against a vertical wall like this with the force F. What are the forces do I have? So, the first one that I usually draw is mg, because there's always mg, its mass times gravity, so it's 10 times 10, IÕm going to round gravity to 10 just to make it easier, and this will be 100. Now, if I push with an F there's a counter force here, that force is called normal, and since I have coefficient of friction, so this is a rough surface, it's not smooth and I have a normal, I don't have friction. Remember, friction is Mu normal and I have both of these gaps, but I want this, I want to know what is the force that I need, right? So that the block does not move, if the block doesn't move, right? Usually it would go down, so if it's not moving it's because friction is opposing that attempt of motion, friction this way, it's a static friction because we want to block not to move, and if it doesn't move, it's because these forces are exactly the same, so the acceleration, the Y axis is 0, so let's go set up. Set up here beginning of this answer is that static friction must equal mg which equals 100, alright? Now, I can expand static friction, static friction is Mu static, normal equals 100. I know Mu static, Mu static is 0.6, so I can solve for normal, normal equals 100 divided by 0.6, I have this here this is 167 Newtons. Now, if you notice, actually normal is the same as the same magnitude as F. F equals normal, because of action-reaction, alright? So, whether you maybe knew this was going to happen already or maybe you stumble upon it and then later realize that F equals normal, but either way, at this point we already have the answer to part A, all right? So all I had to do is find normal because normal is or has the same magnitude as F. So, part B here continuing, says you want the block to begin sliding down the wall, so you temporarily push against with half the value found in part A, what acceleration will in half? So, the idea is that the force that you're going to use, the part-, IÕm going to call this FB because it's part B, is going to be half of the force in Part A, so it's 167 divided by 2 and that is a 83.3, ok? Very similar set up, I got a 10, but now I'm pushing with a force action, know what force it is 83.3, which means there's a normal pushing back that is also 83.3. What else? Got the wall over here, got the wall here and there's an mg pulling down, mg is 100 as well and I have a friction going up. Now, I'm pushing, I needed 167 to keep it from moving, so if I push it with 83.3, that's not enough to keep it from moving, so I know that it's going to move, I don't have to check that, I know that because I'm below what I needed, right? So, this thing will move or more precisely it will accelerate down, so it's accelerating, it's picking up speed, so this friction here is actually kinetic, okay? And what we want to know is what is the magnitude of this acceleration here in the Y axis, okay? How do I find the acceleration on a force problem? F equals ma, sum of all forces in the Y axis is ma y. The forces are: I have fK, I'm going to say that this is positive because it's up and mg down. Mass is a 10 and ay is what I'm looking for. Friction kinetic, let me calculate that here, friction kinetic is Mu kinetic normal, Mu kinetic is 0.4 and normal in this case is 83.3, so friction kinetic, I have it here somewhere, is 33.3, okay? So that I can plug in here, I have 33.3 minus 100 equals 10 ay, and if I move things around I get that ay equals negative 6.67 meters per second squared, okay? That's it for this part. Now part C, so negative, it's negative because it's going down, just to be double sure I'm gonna, just to be super clear I'm going to put an arrow here to indicate that as well, alright? So, part C says shortly after the block begins sliding down you want to keep it moving but with the constant speed, right? To keep change your mind, so you want to keep moving with the constant speed. Moving with the constant speed means that you want your acceleration, the Y axis to be 0. So, the first part I didn't move at all, second part it accelerated down, in the third part I wanted to keep moving but with an acceleration of 0. If I want an acceleration of 0 while it's moving, it means that I want my mg of 100 to be exactly matched by my kinetic friction, so the kinetic friction has to be 100 as well, the walls over here, okay? So, again I want these forces to be the same, so that they cancel, but this friction here will be kinetic because, even though there's no acceleration, this block is moving down, okay? So the set up here, much like how here to set up with mg equals Fs, here I want mg to equal fk. Friction kinetic is Mu kinetic normal, equals 100. And the question here is how much force must you apply? Okay? So this is similar on to part A. Remember, F and N have the same magnitude, right? Action-reaction, so when I find N, what I'm really doing is finding F indirectly, right? So Mu k is 0.4 normal 100, so normal is 100 divided by 0.4, normal is 250, that means that that is your F as well, okay? So it takes a force of 250 to keep this thing from accelerating, but it's still going to fall at a constant rate, alright? If you push any harder, your friction will be bigger than your mg, and this thing would start stopping, and that's not only 1, it would start slowing down, okay? So I have to practice problems I want you guys to try and let's go into that right now.

Practice: The system below does not move. Find the minimum value that μ,S (between 8 kg block and table) can have.

Practice: A 15 kg block is initially at rest on a horizontal surface. The coefficients of friction between the block and the surface are 0.5 and 0.7. How hard must you push down the block to keep a 300 N force in the +x from moving it?

0 of 3 completed

Consider the figure below, graphing the force applied on a 200 kg box vs time. During the first 0.5 s on the graph, the box remains at rest, while during the remaining time on the graph, the box moves at a constant speed of 15 m/s. What is the maximum coefficient of kinetic friction during the motion of the box?

Consider the figure below, graphing the force applied on a 200 kg box vs time. During the first 0.5 s on the graph, the box remains at rest, while during the remaining time on the graph, the box moves at a constant speed of 15 m/s. Assume that the decrease in the coefficient of kinetic friction is due to changes in the surface the box moves across. How far across this surface does the box have to travel for the coefficient of kinetic friction to reach half its maximum value?

A 5 kg box sits atop a 20 kg box inside an elevator. If the elevator was accelerating upwards at 2 m/s2, how much force would you have to push the 20 kg box so that the 5 kg box falls off? Assume that there is no friction between the 20 kg box and the floor, and μs = 0.4 between the 5 kg and 10 kg box.

Consider the figure below, graphing the force applied on a 200 kg box vs time. During the first 0.5 s on the graph, the box remains at rest, while during the remaining time on the graph, the box moves at a constant speed of 15 m/s. What is the coefficient of static friction?

Consider the figure below, graphing the force applied on a 200 kg box vs time. During the first 0.5 s on the graph, the box remains at rest, while during the remaining time on the graph, the box moves at a constant speed of 15 m/s. At what time would the force of static friction be 0.5 kN?

Rank the crates on the basis of the static frictional force acting on them. Rank from largest to smallest. To rank items as equivalent, overlap them. Assume the applied force is the maximum that can be applied without the crate moving.

A 5000 kg truck is parked on a 7.0° slope. How big is the friction force on the truck?Express your answer to two significant figures and include the appropriate units.fs =

Boxes A and B in FIGURE Q6.13 both remain at test. Is the friction force on A larger than, smaller than, or equal to the friction force on B? Explain.

Part A.) Six boxes held at rest against identical walls.Rank the boxes on the basis of the magnitude of the normal force acting on them.Rank from largest to smallest. To rank items as equivalent, overlap them.1.)130N--->7kg2.)150N--->1kg3.)150N--->7kg4.)120N--->3kg5.)140N--->5kg6.)140N--->3kg(Since the boxes are at rest, Newton's 2nd law dictates that the horizontal forces on each box must add up to zero. You can use this information to determine the normal forces. If two boxes are both pushed against the wall by the same force, then they should experience the same normal force.)Part B.) Rank the boxes on the basis of the frictional force acting on them.Rank from largest to smallest. To rank items as equivalent, overlap them.1.)130N--->7kg2.)150N--->1kg3.)150N--->7kg4.)120N--->3kg5.)140N--->5kg6.)140N--->3kg

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