A 9.00-m-long uniform beam is hinged to a vertical wall and held horizontally by a 5.00-m-long cable attached to the wall 4.00 m above the hinge (Fig. E11.17). The metal of this cable has a test strength of 1.00 kN, which means that it will break if the tension in it exceeds that amount.
(b) What is the heaviest beam that the cable can support in this configuration?

Question

A 9.00-m-long uniform beam is hinged to a vertical wall and held horizontally by a 5.00-m-long cable attached to the wall 4.00 m above the hinge (Fig. E11.17). The metal of this cable has a test strength of 1.00 kN, which means that it will break if the tension in it exceeds that amount.

(b) What is the heaviest beam that the cable can support in this configuration?

Diagram showing a beam hinged to a wall, supported by a rope, illustrating rotational equilibrium concepts.

(b) What is the heaviest beam that the cable can support in this configuration?

Accepted Answer

Everyone today, we are going to determine the greatest weight of the window that the rope can withstand without breaking. And I am just going to start with creating a list of givens. So, first we know that the length of the window here, given is 4.5 m long. So that's going to just be um l window equals to 4.5 m. And then next we know that it is captain horizontal position using a seven m long rope. So that is the L. Of the rope, which is going to be seven m. And then the rope hooks six m above the hinges. So that is going to be the distance of between the rope and hinge. I'm just going to write it down as L wall, which is going to just be six m. Okay, so next we know that the maximum load rope can withstand is going to be 0.8 kg newton. And that is just essentially the maximum tension or B. That the rope can withstand 0.8 kg newton. And in S. I. Unit that is just going to be 800 newton. Multiplied by, multiply that by 1000. Okay, so now we know that this diagram here is a bit confusing. So I'm just gonna draw much better representing free body diagram for you guys here. Okay, so Here we have the rope and then we have the wall and also we will have the window here. Okay, so we know that the window length is 4.5 m and assuming that the window is uniform and it will have the center of mass exactly in the middle or center of gravity exactly in the middle which is just going to be the weight of the window right? Which is the one that we want to find. So because of that assumption we know that this X. Right here or this distance is going to be 2.25 m which is just going to be half of the overall length of the window. Okay so next other than this weight right here we know that there are other forces acting upon our system. Right? So first we will obviously have the tension of the rope which is known which is T. And as usual on this sort of problem we want to just immediately do the projection. So the tea or the vertical direction of the tea is then going to just be t multiplied by sine of data. This angle right here is just gonna be data. And then the first the horizontal direction is going to be t multiplied by co sign of data. Okay. Like so so now that we've projected the piece, that should be good. The last force that we have here is the forces of the hinges. So we will have two different forces or two different component. One is going to be on the vertical direction and one is going to be on the horizontal direction which H Y and H X. Okay, I am just going to complete this right here with this is going to be seven m and this is going to be six m okay, like, so, so now that we know all the dimensions, there's one still missing length that we need to know in order for us to actually solve this problem, which is the length from the hinges to the point where the rope and the window actually meet, which is just going to be an X here. So how do we want to find this? X. So we can actually use the pythagorean theory um because we know that this angle right here between the wall and the window is perpendicular or 90 degrees. So we can use the pythagorean theory which is just a square plus B squared equals c squared, with c squared being the hypothesis or the slope. So a squared is going to be the wall square which is six squared plus b squared is going to be our X X squared equal c squared, which is going to be seven squared, So X is then going to be the square root of seven squared minus six squared, which is actually going to be 3.61 m like. So Okay, so we know that the X is 3.61 m. So we know that our free body diagram here is a good representation of our problem earlier, we do not know whether the weight here is going to be before or after. The point where the rope meets the window. But now that we find The distance from the Hinge to that point, which is 3.61, we know that the meeting point here is going to be after 2.25 m or after the weight. Okay, so we know every dimensions here except this angle that we have here. So how do we want to find that angle? We can actually utilize trigonometry, which is pretty much just the sign. Are the tangent of this angle. We can actually use our point here that we just found our our distance which is an X 3.61 m. But I am just going to use the six and the seven which are given from our question. So Because we're using whatever is in front over the hypothalamus, we are going to be using sign of data. Sign of data is going to be 6m over seven m. And recall to find this data, we have to find the inverse of sin and recall that the inverse of sin is going to be arson. So the theater is going to be the arc sine of 6/7 and that all turns out to be 59 degrees. Okay, so now that we solve our dimensions and write everything into our free body diagram here. What we want to do next is to actually start solving the problem. So looking at our system here, we have a lot of unknown forces. We have the H Y, H X, T, w, R T. Is known. So H Y, H X and W. So the way we want to actually tackle this is to use the second equilibrium condition which is sigma torque going to equals two second sigma torque at a certain access C will equals to zero. Because this means that we are going to be able to neglect the forces at our pivot point, which in this case we want to know, we want to indicate that our pivot point is going to be a church Y and H X. So using the this pivot point we can eliminate H Y and H X and therefore we will only have one unknown which is W and neither force which is the which is actually known. Okay, So recall that the torque itself is going to be a multiplication of distance, multiplied by a force acting upon it multiplied by a correction angle. However, if the angle it's if the torque, if the force and their distance is already perpendicular to one another than this correction angle can just be neglected or it's going to be 90°, which is sine of 90°, it's going to be one. Okay, so now recall that using this torque or calculating this sigma torque, we usually use a convention where everything going counterclockwise is going to be in a plus in a positive direction. Well everything going clockwise is going to be in the negative direction. However, just because this is at equilibrium or equals to zero, we can actually do either one. Okay, so I am just going to pretty much use this convention here and start with the first force that we have using this pivot point, which is going to be T multiplied by signed data. So this is going to be T multiplied by signed data, multiplied by our distance, which is going to be X. And this is going to be minus. I'm just gonna put it in parentheses today, it's easier. This is going to be minus the weight, multiplied by the distance, which is going to be our window divided by two. Like so and this will equals to zero because it is at equilibrium. Okay, now we can actually start putting everything that we know into this, into this equation here. So we know that this is this T is going to be 800. Newton Multiplied by sign of 59, Multiplied by 2.61 m minus W, which is unknown, multiplied by 2.25 m. And we know that the W is already perpendicular to our window here, which is why we don't need correction angle here. Right, so now we can actually simplify this and this will just be 2.25 m times W equals the right side, which is going to be 2476 Newton times meter which this will actually then we can just defied this W. Here and find this W. Here 24768, 10 times meter over 2.25 m And then the W is going to be 1100 Newton. So that will be the answer to this particular practice problem. If you guys are still confused, feel free to open the lesson plan on this particular problem. But 1100 Newton is going to be our answer, which is going to be option C. Thank you.

Verified Solution

Key Concepts

Rotational Equilibrium

Tension in a Cable

Torque Calculation