One end of a uniform 4.00 -m-long rod of weight Fg is supported by a cable. The other end rests

step 1 Chapter 12, Problem 23. In this problem, we have a rod suspended by a cable at one end with only

One end of a uniform 4.00 -m-long rod of weight Fg is...

Key Concepts

Static Equilibrium

Static equilibrium is the condition in which a system is at rest, meaning that the vector sum of all forces and the sum of all torques (moments) acting on the system must be zero. This principle is essential for analyzing systems where multiple forces, including gravity, tension, and friction, interact, ensuring that the object remains stationary and balanced.

Torque (Moment of Force)

Torque is the measure of the tendency of a force to rotate an object about a pivot point or axis. In equilibrium problems, calculating torque involves multiplying the force by its lever arm, the perpendicular distance from the pivot. It is crucial to ensure that the sum of the torques due to various forces cancels out for the system to be in rotational equilibrium.

Static Friction

Static friction is the force that resists the initiation of sliding motion between two surfaces in contact. Its maximum value is proportional to the normal force, governed by the coefficient of static friction. In stability problems, static friction plays a key role by providing a resistive force that helps to prevent slipping until a certain limit is reached.

Uniform Rod

A uniform rod has its mass evenly distributed along its length, causing its weight to act at its geometric center, or center of mass. This simplification is important in calculations involving the distribution of forces and torques, as it allows the weight of the rod to be treated as if it were concentrated at one point.

Lever Arm

The lever arm is the perpendicular distance between the line of action of a force and the pivot point or axis of rotation. It is a key factor in determining the magnitude of the torque produced by a force. Understanding lever arms is essential in problems involving rotational equilibrium, as changes in the lever arm length directly affect the calculated moment.

Transcript

00:01 Chapter 12, problem 23.

00:04 In this problem, we have a rod suspended by a cable at one end with only friction forces holding the other end against a wall.

00:11 We need to find the closest position we can place an additional weight without causing the rod to slip.

00:17 First, we have the givens.

00:20 Let's see, we have a 4 meter rod.

00:25 The cable suspending that rod is at 37 degrees.

00:30 We have the mass of the rod providing a weight of f sub g, and that's going to be halfway across the rod, 2 meters.

00:40 And then we have a mystery weight, or, well, we know the weight is the same weight as the rod itself.

00:47 So it's also f sub g, but it's a mystery distance from the point on the wall that they've designated as a, where the rod is resting held only by friction forces.

00:59 In that distance x while it's shown in the illustration as being on the left side of the center of gravity it doesn't mean it's going to actually end up there so don't be fooled by that we're given a coefficient of static friction of the wall of 0 .5 or 1 half so we need to find the forces involved we'll start with a we'll have the the tension force holding it that the cable uses hold it up and over and that's going to have x and y components that we'll be needing here in a minute um so we'll call this t a t for tension and then this is the t y and this is t x and then we're going to have the blank we're going to have the forces actually holding or from the wall against the rod um the reaction reaction force y will be holding, the one that's holding it up, and the reaction force x is going to be pressing it against the wall.

02:17 So now that we have that, we can, by inspection, we can say we can, it's clear that the sum of the forces in the x direction that tx is going to equal the reaction force x.

02:40 Of trig, we can actually see that t of x is going to be t of y divided by the tangent of 37 degrees.

03:01 We won't need that for a little while, but we will end up using that later on.

03:05 So, and in case you're not clear on that, tangent is going to be, or the tangent of the angle is going to be the opposite.

03:15 Over the adjacent or ty over tx.

03:17 So we're basically just doing the algebra to convert that to ty over tangent of 37 degrees as tx.

03:28 All right.

03:29 The next thing we can see by inspection is that ry is going to be the coefficient of static friction times the normal force.

03:40 But in this case, we've already been given a static coefficient of static friction with one half.

03:47 And we know that's going to be, and the normal force is going to be tx.

03:54 And then just to take it a step further, we'll go ahead and point out that ry is going to end up equaling ty over 2 times tangent of 37 degrees.

04:19 I didn't specify meters, did i? all right.

04:23 So much for setup.

04:25 The next step is to start doing moments.

04:29 We'll start and we'll declare counterclockwise as positive with a sum of moments.

04:40 We're looking for ty, so we're going to start with a sum of moments around a.

04:47 I suppose we could do it in the other direction.

04:49 That's not how i did it, so i'm going to stick with what i've already written down.

04:55 All right, so again, since we got positive as counterclockwise, the moment from fg of our mystery weight is going to be fg times 8.

05:12 The moment of the mass of the bar is also going to be negative counterclockwise.

05:22 So we'll go ahead and call that f of g times two.

05:26 Because it's two meters over.

05:30 And then finally, we have a positive moment from ty times four.

05:45 And so we end up, if we group our term, we subtract fgx and fg2, or add fgx and fg2 to both sides, end up the t, y times, oh, let's go ahead.

06:00 Yeah, we'll go ahead and do this.

06:02 Times 4 is equal to.

06:08 And we're going to go ahead and group the fg and just have x plus 2.

06:21 And then we'll go ahead and subtract or subtract, divide both sides by four.

06:27 And so ty is going to end up equalling f of g times x plus 2 divided by four.

06:35 All right, so we can now go back up here where we established ry and find that, you know, what, let's wait until we've done the next moment calculation.

06:59 We'll hold off on that...

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