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A uniform beam having a mass of 28 kg and a length of 5.9 m is held in place at its lower end by a pin: Its upper end leans against a vertical frictionless wall whe...

Question

A uniform beam having a mass of 28 kg and a length of 5.9 m is held in place at its lower end by a pin: Its upper end leans against a vertical frictionless wall where the angle between the beam and horizontal surface is 51 degrees as shown in the figure: What is the magnitude of the force the pin exerts on thebeam?680 N460 N740 N520 N300 N

A uniform beam having a mass of 28 kg and a length of 5.9 m is held in place at its lower end by a pin: Its upper end leans against a vertical frictionless wall where the angle between the beam and horizontal surface is 51 degrees as shown in the figure: What is the magnitude of the force the pin exerts on the beam? 680 N 460 N 740 N 520 N 300 N



Answers

A uniform beam is 5.0 $\mathrm{m}$ long and has a mass of 53 $\mathrm{kg} .$ In Fig. $12-$ $74,$ the beam is supported in a horizontal position by a hinge and a cable, with angle $\theta=60^{\circ} .$ In unit-vector notation, what is the force on the beam from the hinge?

In this problem. On the topic of equilibrium and elasticity, we have shown a uniform beam which has a weight of 60 newtons and a length of 3.2 m. It is hinged at its lower end and a horizontal force of magnitude five newtons acts the up end. The beam is held vertical by a cable, making 25 degrees with the ground and is attached to the beam at a height of two m. We want to calculate firstly the tension in the cable and next the force on the beam from the hinge. Now, if we choose an access through the hinge perpendicular to the plane of the figure and take talks, that would cause a clockwise rotation to be positive. Then we require the net talk to vanish since the beam is in equilibrium and this net talk fl sign of 90 degrees minus T. H. Sign of 65 degrees must equal to zero here the length of the beam L is given to be 3.2 m. And the height at which the cable attach is is H. And age is attached at two m. And we note that the weight of the beam does not enter in this equation, since this line of action is directed towards the hinge. So if we take if as 50 newtons has given in the problem statement, we can rearrange the equation above and solve for T. And we get T. Is equal to F times L over H. Sign of 65 degrees. If we put our values into this, we get this to be 50 newtons times three 0.2 m, divided by the height two m times the sine of 65 degrees. This gives us detention T. To be 88 newton's and that was the solution for part A. In part B. We want to find the force on the beam from the hinge in unit vector notation. So to find the components of the force F. P. Will call it. We have to balance the X and Y components, so we know that some of the X components of this force was equal to zero, which means if P X is equal to T. Timely co sign of 25 degrees minus Mhm Force F, which is purely horizontal. Similarly, we have these some the vertical forces was also equal to zero for equilibrium. In this case, that's if P Y must equal to T. Sign 25 degrees plus the weight of the rod W. And that weight of the beam is 60 newtons. So we therefore find that the hinge force components, when we calculate uh F. P. X, which is 30 newton's, and this points to the right. And if P Y is 97 newtons pointing upward. So when you infect a notation, we can write this force that the hinge applies to the beam. F p is simply 30 newton's long, unit factor i less 97 newtons times unit vector J.

So here. Here's a free body diagram of the system and we know that the position of the center of mass from the success suspension point is going to be denoted as s. And this week will one meter. So we can say that for party, the network is going to be equal to zero because this is in rotational equilibrium and this would be equal to the weight times over to the length plus s and this time sign of fate A. And then we have plus the weight terms s time sign of Fada and then minus Ah, the weight Several one. Let's do a new line Now that we can say plus think waits times asked time sign of data. I'm in here minus the weights of one times all over too. Minus s sign of Fada. So we know that sign of fate is going to cancel out and we have that for W plus the weight minus two times waits up. One. This will equal zero. So we find that weight is going to be equal to two times the weights of one minus w divided by four. And so this will equal two times 100 minus 1 40 divided by four. And we know that weight. The lower case w. Was going to be equal to 15 Nunes and then it's asking us what happens of theta changed. So what happens if this angle right here where to change and well, we know that the final result? No, some hard, eh is independence of thie angle measure. So we say that the wait would stay constant at 15 meters. 15 Nunes Rather my apologies. That is the end of the solution. Thank you for watching.

Left. We have the free body diagram and we can then fined for these. We know that this system is under static equilibrium. So we can say that the net torque for the sum of the torques is gonna equal zero. And this is gonna be equaling the force multiplied by zero because we are evaluating at this origin point. So let's just say that some of towards that the origin and so this will be plus mg multiplied by the length over too, minus the tension force multiplied by length minus D. And so we can then say that four part of the tension in the rope must then be equaling two t equaling mg multiplied by L over two, divided by L minus D. And so this would be equaling 343 new tins, which would be the weight multiplied by 2.50 meters. This would be divided by 5.0 meters minus 1.20 meters, and we find that the tension must be equaling it. Ah, 226 new tents approximately. This would be our final answer. For part B. We condense. Find that the due to Newton's second law for Part B. Applying Newton's second law on the Y direction. This is equaling zero. This is equaling the force, plus t minus M. G. We're trying to find f. Of course, the force that the column is it is exerting. And so we can see that then the forces equaling mg minus t. Because, of course, the tension in the rope is helping. Other are decreasing the magnitude that the magnitude of the force that the column is, um, is exerting. And so this would be equaling. 2 343 mutants minus 226 new tunes that we found in part day. So this is equaling 117 Newton's upward. This would be our final answer for Part B. That is the end of the solution. Thank you for watching

All right. So this problem is asking us to determine the initial angular acceleration and the magnitude of the force at point A which is a pin. And so the first thing we want to do is find inertia, which is for this problem one third m elsewhere. And this is a moment of inertia for a beam that's pinned through its end. Pretty much every shape is going to have its own moment of inertia. So you always want to be sure that you're using the right one when you work a problem. So for example, if this wasn't pinned at the end, but it was pinned in the center, that would change the moment of inertia and this would be 1/12 and elsewhere instead, but it's pinned at the end. So we have the right moment of inertia And we know em and we know LM is 100 and LS four. So we can go ahead and plug those in. We can get 500 and 33 .3 reputation for Inertia. And now we can move to our net next step, which is to some the moments to find alpha. And so We can some P which is the force that's pulling at the bottom, which is equal to 300 newtons times sine of theta where theta is equal to 45 degrees. And that's also given in the problem Times three Because that's a distance the way it is. And we divide by inertia to get a expression for alpha. So we can solve for alpha if we plug all of those aim And we get that alpha is one one 93 radiance per second squared. All right. So now we can find acceleration in the X direction using A X equals R alpha. And so our is to for this problem, so we're just gonna double alpha. And that's going to become too 0.3 86 And we also know that A. Y. Is going to equal zero, since the angular velocity of the beam Is uh is also zero, there's no angular velocity. Yeah. And so now all we gotta do is some of the forces. And we're going to start with the horizontal forces. So A X. Which if you remember from our free bio diagram, is the force at this pin in the horizontal direction, That's going to be added with P sign of 45, which is data equals M A X. And so now we know A X. You know, em we know P. And so we can go ahead and plug all of those in And we're going to get 26 0.5 newtons. And so next we do the same thing, but for the vertical direction, so we have a Y plus P. Co sign of 45 minus MG. Since gravity again from the free bio diagram is going downwards and we already checked the A. Y. Is zero. So we can go ahead and cancel this term. We add MG to the other side. We subtract PICO sign 45 And we get that A.Y. is equal to 700 And 69 Nunes. So the final step to find the result in force. We just use this formula here, F. A. That's our result, enforce. We square the horizontal and the vertical components A. X. And A. Y. And then we take the square root of that. And so this will give us 700 and 69 0.5 newtons. And so that's our result in force.


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