A 15,000-N crane pivots around a friction-free axle at its base a...

Question

A 15,000-N crane pivots around a friction-free axle at its base and is supported by a cable making a 25° angle with the crane (Fig. E11.18). The crane is 16 m long and is not uniform, its center of gravity being 7.0 m from the axle as measured along the crane. The cable is attached 3.0 m from the upper end of the crane.

When the crane is raised to 55° above the horizontal hold-ing an 11,000-N pallet of bricks by a 2.2-m, very light cord, find (a) the tension in the cable. Start with a free-body diagram of the crane.

Accepted Answer

Hey, everyone in this problem, we have cranes that lift objects using a boom. A 10,000 Newton boom is pivoted about a frictionless axle at its lower end, A cable attached at five m from the upper end making an angle of 18° is used to support the boom. The boom is 25 m long and non regular. The centre of gravity is located 11 m from the axle measured parallel to the boom. The boom is directed at 60 degrees above the horizontal, suspending a 45 100 Newton container using a 1. m long massless rope. And we're asked to find the tension in the cable. The answer choices are a 1.5 times 10 to the four Newtons. B 1.9 times 10 to the three newtons C 5.6 times 10 to the three newtons, the 1.7 times 10 to the four newtons and E 1.8 times 10 to the four newtons. Now, we're given a diagram here. Let's convert this into a free body diagram. So we're gonna have this horizontal line. Can you draw it out like so that the boom connects to. And then we have our boom. We know that the angle between the boom and this horizontal is 60°. That's in our diagram. And we're also told in the problem That the boom is directed 60° above the horizontal. We have our tension from our cable as we have our cable here. And it has a tension force. We can call it t The angle between the boom and that forces 18°. And what other forces do we have? Well, we have this container so we know we have the weight of the container acting downwards from the end of this boom. We're gonna call that W C. We also have the weight of the boom acting downwards. So we're gonna call that w now the distance here between that weight And the pivot at the end is 11 m okay because that's where the center of gravity is. And so that's where the weight is going to act downwards. And then we may have some force acting between the top of the crane in this boom, okay in the left hand direction. And we're gonna call it. And the reason we have it pointing to the left is we have tension in the X direction. If you imagine breaking T into its X component, it would point to the right and we have an equilibrium situation. And so we expect that some of the forces to be equal to zero. So if we have a an X force acting to the right, then we expect another one to be acting to the left to kind of cancel each other. Okay. And so we expect this force would be acting to the left. Now where has to find the tension t what do we know? Well, we have equilibrium condition. So in an equilibrium situation, we know that the sum of the forces is equal to zero. And like I just mentioned in this case, we have forces in the extraction and we have forces acting in the Y direction. And so we have that the sum of the forces in the X direction is zero, the sum of the forces in the Y direction is also equal to. And we also have that the sum of the torques Is equal to zero. Well, which equation should we use? Well, we have this tension T that we don't know and we're looking for, but we also have this force F that's unknown. Okay. So if we start using our sum of forces equations, we are gonna end up with an equation that has both F and T in it, it's gonna be not possible to solve for T without finding F. If we use our torque equation though, we can choose the pivot of the boo Okay down here where it meets the top of the crane, we can use that point as our pivot for our torque. And that means that this force acting right at that point F will not contribute to the torque. So it won't be in the equation. And the only unknown will have is teeth. So let's do that. Instead, let's work with our some of torque equation. So we have that the sum of the torque is equal to zero. Now, what torques do we have to consider? Well, we're going to consider the torque due to the weight of the container. So we have the torque caused by the container. Now, we can imagine this weight would push the boom downwards in the counterclockwise direction. Okay. So we have this counterclockwise rotation which indicates a positive torque. Okay, then we have the weight in the middle here, the weight of the boom, it's also acting downwards. So it would also cause counterclockwise rotation which would give us a positive torque. So we get the torque from the boom is positive and we have our attention acting kind of down into the right. We can imagine that this would actually cause clockwise rotation to push it like this. If it's causing clockwise rotation, that's going to be a negative torque. So we have minus the torque due to the tension T And this is equal to zero. Alright, so we have three torques to consider. Now recall that the torque is going to be the distance from the pivot to the force times the magnitude of the force times the angle between the force and the pivot access. So let's start by finding these ankles. If this angle is 60° down here between the horizontal and the boom, then the angle between the weight of the boom acting downwards and the boom itself, it's going to be 30° They have to sum to 90. And then the angle up here between the weight of the crate and the boom is also going to be 30 because these are peril. Alright. So back to our torque equation, we have the torque due to the crate. So that's going to be the distance between that force, the weight of the crate and our pivot point at the bottom of the boom. Okay. Now, this boom is 25m long. And so that's going to be a distance of 25 m time's up force which is W C, the weight of the crate Times the sine of the angle. So sign and the angle we found was 30°. Okay, then we have the torque due to the boom. Okay. So the distance we're told that the center of gravity is 11 m from this axis or this pivot at the bottom of the boom, that's where that weight is going to act at the center of gravity. And so the distance is 11 m we multiply that by the force which is the weight of the boom, W times sine of the angle until we have signed of 30 degrees. And finally, we subtract the torque due to the tension T so tension T we are told that this cable is five m from the top of the boom. Okay. So the distance between the top of the boom in this cable is five m, which means that the distance from this cable down to the bottom of the boom is 20 m. Okay. The entire boom is 25 m, 25 minus five m gives us 20 m. That's going to be 20 m. The forces attention T and the angle we're told is 18°. So we get signed of 18° and all of this is equal to zero. Okay. I've just written it down below because we've run out of now. We want to isolate for T. So let's go ahead and rearrange and do that. So moving the term with attention to one side, we have 20 m time signed of 18° times attention T is equal to 25 m. Now the weight of the container we're told is 4500 newtons. So we can substitute that in. So we have 25 m times 4500 not meters, 4500 newtons times sine of degrees Plus 11 m. The weight of the boom we're told is 10,000 newtons, Mr 0 10,000 newtons times sine of the angle which is 30 degrees. And so our attention T is going to be equal to everything on the right hand side here, 25 m times 4500 newtons Times sine of 30° plus 11 m times 10, newtons, times sine of 30 degrees, all divided by 20 m time sign of 18°. And when we work all this out on our calculator, We get a value of 18, newtons. And we saw our answer choices are written in scientific notation, Two significant digits. So this is going to be equal to 1.8 Okay. And we have 1234 Times 10 to the four Newtons. And so that is our answer. That is the tension force on that cable. And if we look at our answer choices, okay, that is going to correspond with the answer choice. E the tension in the cable is 1.8 times 10 to the four newtons. Thanks everyone for watching. I hope this video helped see you in the next one.

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