A large wrecking ball is held in place by two light steel cables (Fig. E5.6). If the mass m of the wrecking ball is 3620 kg, what are (a) the tension TB in the cable that makes an angle of 40° with the vertical and (b) the tension TA in the horizontal cable?

Question

Diagram showing a wrecking ball held by two cables, with angles and tensions labeled.

Accepted Answer

Hi, everyone. Let's get started. So a container being lowered from a crane is held in place by a light steel cable that makes an angle of 58 degrees from the vertical and a horizontal synthetic cable. If the container has a massive m equals 45 20 kg, determine the tension T one and the slanted cable and determine detention T two in the horizontal cable. So first, let's figure out what we're given. We have a mass 4520 kg and were given an angle of 58.0°. And now we need to figure out what we need. We have, we need attention, T one and 2. But in order to find those who are going to need to split them into their X and Y components. So in red here, I'll draw the forces on the Y axis And we're going to have T1 y and we also have gravity pulling down in the Y axis, that's a G and then the X axis will complete our right triangle For tension one. And we also have just in the X-axis, our attention force to because it's horizontal. So Now, because we have a right triangle, we can actually figure out that we have a 90° angle here. And if you subtract those from 180, you'll find that this one is 32° in the corner. Now, let's because we know our gravitational force here, we can apply Newton's second law to solve for our other force in the Y axis. So we can take our equate our new and second law equation F equals M A make some substitutions. So we have two forces In our, some, we have our upward force, T one Y minus our downward force of MG. And that will be equal to zero because this is a tension force. We know there's no acceleration going on. So now we can rearrange that To have T- one y equal to MG. And now we can actually, because we know M G we can solve for it MG equals 45, 20 kg Times 9.8 m/s square. That's our gravity. And that's going to give us 44,296 newtons. That's our gravitational force. So now that we know that they're equal to each other, we can also say that T one Y Is equal to 44,296 Newtons. Now, we can use some trick here too, Figure out our attention force T one which is the hypotenuse. And if you recall the tension force in the Y axis is equal to the total tension force times the sine of our angle right here, measured from the vertical 32 degrees. And then when you solve that, you'll get a total tension force T one of eight point 36 times 10 to the four newtons. Now, let's see if we can do some analysis on our multiple choice problems or answers rather to determine if we can narrow anything down with process of elimination. So we're looking at our T ones here because that's what we just saw for. We got an 8.36 times 10 to the four. So let's see, this one's 4.43, that's not gonna work. This one's 5.62, that's not gonna work either. This one's also 4.43. So that's not gonna work. And then answer D actually does have 8.36 times 10 to the four. So that must be the correct answer. D T one equals 8.36 times 10 to the four Newtons and T two equals 7.9 times 10 to the four Newtons. Thank you.

A large wrecking ball is held in place by two light steel cables (Fig. E5.6). If the mass m of the wrecking ball is 3620 kg, what are (a) the tension TB in the cable that makes an angle of 40° with the vertical and (b) the tension TA in the horizontal cable?

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Key Concepts

Tension in Cables

Equilibrium of Forces

Trigonometric Functions