Skip to main content
Ch 05: Applying Newton's Laws

Chapter 5, Problem 5

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?

Verified Solution
Video duration:
4m
This video solution was recommended by our tutors as helpful for the problem above.
3074
views
Was this helpful?

Video transcript

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.
Related Practice
Textbook Question
A 45.0-kg crate of tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it, and the crate just begins to move when your force exceeds 313 N. Then you must reduce your push to 208 N to keep it moving at a steady 25.0 cm/s. (a) What are the coefficients of static and kinetic friction between the crate and the floor?
2019
views
Textbook Question
A 45.0-kg crate of tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it, and the crate just begins to move when your force exceeds 313 N. Then you must reduce your push to 208 N to keep it moving at a steady 25.0 cm/s. (b) What push must you exert to give it an acceleration of 1.10 m/s2?
891
views
2
rank
Textbook Question
A 1130-kg car is held in place by a light cable on a very smooth (frictionless) ramp (Fig. E5.8). The cable makes an angle of 31.0° above the surface of the ramp, and the ramp itself rises at 25.0° above the horizontal. (b) Find the tension in the cable.

1670
views
Textbook Question
A box of bananas weighing 40.0 N rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.40, and the coefficient of kinetic friction is 0.20. (a) If no horizontal force is applied to the box and the box is at rest, how large is the friction force exerted on it?
545
views
Textbook Question
A box of bananas weighing 40.0 N rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.40, and the coefficient of kinetic friction is 0.20. (b) What is the magnitude of the friction force if a monkey applies a horizontal force of 6.0 N to the box and the box is initially at rest?
463
views
Textbook Question
A box of bananas weighing 40.0 N rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.40, and the coefficient of kinetic friction is 0.20. (c) What minimum horizontal force must the monkey apply to start the box in motion?
510
views
1
rank