In Fig. 12-47, a non-uniform bar is suspended at rest in a horizontal position by two massless c

In Fig. 12-47, a non-uniform bar is suspended at rest in a...

Key Concepts

Static Equilibrium

Static equilibrium is a fundamental concept in mechanics where an object remains at rest because the net force and net torque acting on it are both zero. This principle is used to analyze systems by setting up equations that balance forces and torques, ensuring that the object does not translate or rotate.

Torque Equilibrium

Torque equilibrium involves ensuring that the sum of all torques (moments) about any pivot point is zero. This condition is used to determine unknown forces or distances in a body, particularly when it is subjected to forces at various positions and orientations.

Center of Mass

The center of mass is the point at which the mass of an object can be considered to be concentrated for the purpose of analyzing translational and rotational motion. In problems with non-uniform distributions, such as a non-uniform bar, finding the center of mass is crucial for understanding how weight is distributed and how it affects torque.

Force Resolution

Force resolution refers to the technique of breaking down forces into components along chosen coordinate directions, typically horizontal and vertical. This method simplifies the analysis of systems when forces act at angles, allowing the separate application of equilibrium conditions in each direction.

Transcript

0:00 Hi there.

00:01 So for this problem we have the situation that is shown in this figure.

00:06 There is a non -uniform bar that is suspended at rest in an horizontal position with two cores.

00:15 Now one course made an angle theta which we have the value of 36 .9 degrees.

00:26 We also have that the outer angle, fee, is 53 .1 degrees.

00:40 And the length of the part is also given for this, and that length is 6 .10 meters.

00:54 Now, what we need to obtain from this is the distance adds from the left end of the bar to the center of mass.

01:04 So we need to determine that distance adds.

01:09 Now, the first thing that we need to do is to draw all of the forces that are acting on this part.

01:20 Now, we have this fort at the right, this tension.

01:29 We are going to call that tension the art, the tension at the right.

01:33 And the other tension is the tension at the left.

01:38 We call that the tension at the left as t -elt.

01:43 And we also have the weight of this part.

01:51 Now, those are all of the forces that are opting on this part.

01:56 Now, with this, we know that the bar is in equilibrium.

02:00 So the forces and the torts acting on it, each sum to zero.

02:08 Now, with that said, we will obtain that.

02:11 The sum of the forces, let's put in here, the sum of the forces in the white component, we will have, the tension.

02:23 We can see from the picture that the ones, all of these contributes to the white component of the forces, we will, and since the angle is we will, with respect to the vertical for both tensions, the components is given by the cosine of that angle.

02:43 So, for example, for the tension on the left, that is the tension on the left, cosine of theta.

02:54 And it is positive because it is pointing upward.

02:58 And we also have the tension at the right, a sign of fee, and the weight that is pointing, so it's negative and this should be equal to zero.

03:12 Now we do something similar, summon the forces in the x component.

03:17 So as you can see from the picture, the only ones that contribute to that are detentions because the weight has no ex component.

03:28 So with that said, we will have the tension at the right sign of theta, and this is positive because it is pointed to the right minus the tension at the left, sign, oh sorry, this is sign of fee, and this is sign of t -town.

03:50 And this is equal to zero.

03:53 And finally, we sum all of the forces that make torque in here.

04:05 And we know that in this case, the only two forces that may torque is the force, is the tension at the right, and the weight because the point of rotation is here at this end, at the left end of the bar.

04:26 So the tension at the left makes no torque.

04:32 So with that set, we will have, if we apply the right -hand ruled, that the torque of the weight is positive...

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