Two identical, uniform beams are symmetrically set up...

Key Concepts

Static Equilibrium

Static equilibrium is the condition in which an object is at rest and all the forces and torques (moments) acting on it balance each other out. This concept is crucial because it ensures that the sum of forces in any direction and the sum of moments about any point are zero, which is necessary for analyzing whether a structure or object will remain in a stable configuration.

Friction

Friction is the resistive force that opposes the relative motion between two contacting surfaces. In static scenarios, this force is typically characterized by the coefficient of static friction, which defines the maximum frictional force before slip occurs. The concept is essential for determining the conditions under which an object can remain at rest without sliding along the surface.

Normal Force

The normal force is the reactive force exerted by a surface perpendicular to the object in contact. This force is a critical component in friction calculations because the frictional force is proportional to the normal force. Understanding how the normal force is distributed and its magnitude is key to predicting the onset of motion or tipping.

Torque (Moment) Equilibrium

Torque, or moment, refers to the rotational effect of a force about a pivot point. When analyzing objects on the verge of tipping or falling, it is important to consider how forces create opposing moments. Achieving torque equilibrium, where all clockwise moments balance the anticlockwise ones, is necessary to ensure that the object remains balanced.

Free-Body Diagrams

Free-body diagrams are graphical representations that isolate an object to depict all the forces acting on it. They help in visualizing the problem and setting up the equilibrium equations by clearly showing forces such as weight, normal forces, frictional forces, and any applied forces. These diagrams are fundamental tools in solving static equilibrium problems.

Center of Mass

The center of mass is the point in an object where its mass is considered to be concentrated. The location of the center of mass affects the stability of objects; if the vertical line through it falls outside the base of support, the object will tip over. This concept is important when analyzing the balance and stability of structures or arrangements.

Transcript

00:02 Okay, so chapter 9, problem 71.

00:04 Two identical uniform beams are symmetrically set up against each other, as in figure 9 .87.

00:10 On a floor with which they have a coefficient of friction, mu s is equal to 0 .50.

00:17 What is the minimum angle the beams can make with the floor and still not fall? right.

00:24 So we want to find the minimum angle that they won't fall at.

00:27 So as always, with a lot of these questions in chapter 9, we want to use a free -border diagram.

00:32 And here's the diagram right at the bottom there.

00:36 The bottom left.

00:39 So, of course, we have the center of mass is where all the masses.

00:42 Well, the mass is going to act as if it's centralized there at the middle.

00:49 And so you're going to have mg pointed downwards there.

00:52 You want to have the force.

00:53 So here i've only drawn the left beam.

00:55 So i'm saying draw a free body diagram of one beam because you can really use the forces.

01:00 And when you understand the forces, you really don't need to.

01:04 Have both of them there.

01:06 I mean you can draw both of you like so feel free but i'm just going to draw one.

01:13 So yes we have the force of the beam at the top right.

01:18 So let's just go from the top and downward.

01:21 From the top of the free biode diagram like northeast to south -west.

01:26 So the force from the beam, that's just from the right beam hanging up next to it right? so the beam is going to push the right beam is going to push it that way and newton's third lob by the way to tells us that if the right beam is pushing on it that way, then there also must be an equal and opposite force on it from the left beam, right? so if the right beam is pushing the left downwards, then the left is going to push the right upwards, right? so that's one thing to note for future.

01:57 I already mentioned the force due to gravity.

02:01 And we also have the angle there at the bottom that will be given.

02:04 Or actually, no, we won't be given.

02:07 I may find this later.

02:09 We're searching for the angle.

02:12 That specific angle, not the one above the line or anything, is just going to be what we're looking for.

02:16 And then you also have the normal force pointed upwards from the ground, so the ground's pushing the pole up, and then you have the friction force for whenever it starts sliding.

02:26 If you can imagine the beam falling to the left, or sorry, falling to the right.

02:31 And so eventually that bottom piece, the junction where the normal force and the friction meet is going to move to the left.

02:38 And as we all know, friction goes opposite of the direction that you move, so that's why it's pointing inwards or to the right.

02:48 And with all of this, we have to be able to work with these angles, so when it comes to the length of the beam.

02:56 Because we can't just call it l, right? it's not pointed straight up.

02:59 So it is dependent on the angle what the length of the beam is, what the x length and the y length will be.

03:06 So in terms of in terms of in terms of y, you're going to have l sine theta as the length and then in x you're going to have l cosine theta as the length as was stated before in the problem, the coefficient of static friction is 0 .50 so we're going to use that later from this we can start doing some newton second law and see what we can get out of that so some of the forces in the y direction you have normal force up gravity down that just means that the normal force is counteracting gravity to to keep it above ground.

03:38 So it's equal to mg.

03:40 If we go to some of the forces in the x direction, now you see that there's only two forces there.

03:46 You have the force due to friction and the force of the other beam.

03:50 And so these both are equaling to zero to some of the forces because it's assumed that in that moment that we're looking at it, it's not moving, but it can move.

04:00 Right, so.

04:04 But this would just tell us that the frictional force is equal to the beam force.

04:08 But we're not really going to focus on that much.

04:11 It's not so important.

04:12 The one that's really more important is below it.

04:15 That generally the frictional force is equal to the coefficient of whatever friction is happening.

04:20 In this case, it's static because we want it to not be moving.

04:24 We want it to get to that minimum angle where it will stop.

04:27 And that's really why we're allowed to say it's zero, so there's no acceleration.

04:33 But it could easily be kinetic as well.

04:35 We could have a coefficient of kinetic friction, coefficient of restitution, there's a lot of different things that could fit in that spot.

04:41 But in this case, it's going to be the force of friction is equal to the coefficient of static friction multiplied by the normal force, which we already know the normal force.

04:51 It's just mg.

04:52 So it's just mg times mu.

04:54 Very, very simple, right? but the next part is going to require us to take the torque about the top of the beam.

05:04 So we can't really use...

05:10 We can't, we are, actually, we already did, right? i was going to say we can't use the forces here, but we already used the forces.

05:17 And we basically got all the information that we could out of newton's second law, at least in terms of, at least in the linear terms.

05:26 But we didn't think about it rotationally because it isn't going to be falling.

05:29 And so there's going to be forces acting at a distance, which if you think about it, cause a torque, right? so if there's an angle between, if you have, have a force acting at a distance on a lever arm, for example, like a wrench, then there's going to be a torque.

05:45 And if you have it, it's going to be maximized whenever the angle between them is 90, so we always want to go for those 90 angle torques.

05:55 But anyway, you want to take the torque about the top of the beam.

05:59 So this is just a portion of the drawing i drew here.

06:03 Just that's the f beam pointing to the left, and that the arrow that i drew is just the direction of positive torque.

06:10 So we're going to call clockwise positive, and we're going to call counterclockwise negative.

06:14 That's what i have in the parentheses there.

06:16 And then the side note, we're going to say that the force due to the beam is strictly horizontal, not only because of symmetric geometry, but that is one reason.

06:26 Since it's symmetric, they're going to cancel each other out, and the vertical forces, i mean, will cancel each other out, and then you're going to have to own deal with horizontal, but also the beam force is strictly horizontal since newton's third law tells us that if the right beam pushes up on the left, the left will push down on the right and so on and so forth.

06:47 This is what i mentioned at the beginning.

06:51 So ignore the crossed -out stuff, but i'll read the rest.

06:53 So we have also we choose force from the beam as the anchor point since no torque will be created as a result.

07:01 So to have a torque, you need to have a distance.

07:03 You need to have a force acting at some distance.

07:06 So if we choose the point where the beam force is, there can't be, if we choose that as our anchor point where it's going to be rotating about, there can't be any distance...

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