Finding the couple vector of this ice boat

In summary, The problem involves analyzing forces on an iceboat using vector addition and cross product. The given figures are inconsistent and it is necessary to choose a specific point to calculate the moment. The author has chosen to place the origin at the end of the left-runner to simplify the calculations. The second figure is technically correct but skips steps in going from the first figure.
  • #1
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Homework Statement


The problem and its solution are attached as TheProblemAndSolution.jpg.

Homework Equations


Vector addition. Cross product.

The Attempt at a Solution


Given that I am just supposed to take Fig. 1 for what it is based on the way the problem is phrased, my biggest confusion is in understand how Fig. 1 was translated to Fig. 2 as well as in interpreting Fig. 2.

Firstly, I feel it shouldn't matter what the orientation of the boat is since the forces we are considering in the diagram will have the same magnitudes and they will all be pointing in the same directions with respect to each other. More specifically, defining the axes to be aligned with the orientation of the boat, the 20 lb force on the sail seems to me like it should be 5 feet upward in the same direction as the i_3 vector while directed in the direction of the i_2 vector but it should not be 6 feet from the tail of the i_2 vector. The other 20 lb force vector's tail seems to me like it should be placed on the top-right corner of the bottom face of the rectangular prism directed toward the opposite direction of the i_2 vector instead of it being translated 6 feet in the direction opposite the i_2 vector from where I feel it shoul dbe. As for the 10 lb force pointing downward, I agree with where it is. For the upward pointing 10 lb force vector, I also feel it should be on the top-right of the bottom face of the rectangular prism but, unlike 20 lb force vector that is not applied to the sail, it should be pointing in the direction of the i_3 vector.

Looking online, I found: (1) “In mechanics, a couple is a system of forces with a resultant (a.k.a. net, or sum) moment but no resultant force.” and (2) “Another term for a couple is a pure moment”.

What I found online makes sense to me so I get what a couple vector is and I also get the algebra in computing it. That is, it's the non-zero vector sum of all the torques (applied per force) where the net force is zero.

To me, the r_n and f_n numbers obtained for the 10 lb and 20 lb forces do not seem consistent with the information provided by Fig. 1.2.

If you need me to say more, just ask.

Any help in solving this problem would be greatly appreciated!
Thanks in advance!
 

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  • #2
The only way to arrive at the solution, in my opinion, is to completely forget about Fig.2. It is, as you pointed out, inconsistent with Fig.1. This problem is so badly written the author ought to be ashamed of him/herself.

Whenever you take the moment of a system of forces, it is always with respect to a specific point in space, we'll call it P. If you have the freedom to choose both P and the origin, then you can choose to make them the same point, P = (0,0,0). This is helpful because if you place P = (0,0,0) on the origin of a force vector, you know that its moment about that point P is zero, because r = (0,0,0).

We seemingly have that freedom in this problem (although if the solution hadn't been provided, there would have been no indication of that). Looking at Fig.1, we can place our origin (0,0,0) and the point P, about which we calculate the moment, at the end of the left-runner, where the 10 and 20-lb forces originate. If we do that, we don't have to worry about their individual moments, because once again r = (0,0,0).

Again, we're going to let our orthonormal basis originate from the end of the left-runner where the 10 and 20-lb forces originate (so that we only have to calculate two individual moments instead of four). Let i1 point toward the end of the iceboat (that is, in the direction of the 20-lb force crossed with the 10-lb force), let i2 point toward the body of the iceboat (that is, in the direction of the 20-lb force on the left-runner), and let i3 point directly upward.

Now, using the basis described above, you should be able to calculate the individual moments for the remaining two forces. Unlike the author, you should get r20-lb = (3,6,5) crossed with f20-lb = (0,-20,0) for the second individual moment. Add the two individual moments together, and you should arrive at the author's answer.

It wasn't at all clear to me why the author chose to take the moment about the end of the left-runner, other than for the sake of reducing the amount of math involved. Perhaps there's another way of reconciling the two figures, or of choosing a basis such that you obtain the same individual moments as the author. But I wouldn't spend too much time trying to decipher the author's cryptic (and likely flawed) thought processes. If this problem is out of a book you're required to use for class, I'd recommend getting a supplementary book. If you'd like suggestions on some good statics texts, just send me a PM. Or if you have more questions about the problem, feel free to ask.
 
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  • #3
The iceboat is a rigid body. O force acting on a rigid body at some particular point may be moved to another point at that same body along the line of its direction. That's what the author has done. The 20 lb force acting on the runner was moved to the zero point to simplify the computations. That's a useful technique, master it.
 
  • #4
Sorry for the late reply. After having read what you guys said and thought about the problem for a long time, I can now say that I fully understand it :) . Thank you.

To add, I can now see for myself that the second figure is technically correct but inconsistent with the first figure. In other words, it seems that steps were skipped in going from the first figure to the second figure.
 
  • #5




Thank you for your detailed explanation of your thought process and confusion regarding this problem. It seems like you have a good understanding of the concept of a couple vector and how to calculate it. However, I agree with your assessment that the placement of the force vectors in Fig. 2 may not be entirely accurate based on the information provided in Fig. 1. It is important to have a clear and accurate representation of the forces and their respective locations in order to accurately calculate the couple vector.

I would suggest carefully reviewing the given information and making sure that the placement of the force vectors in Fig. 2 accurately reflects the information in Fig. 1. It may also be helpful to draw a separate diagram with the correct orientations and distances to better visualize the problem.

Additionally, I would recommend double-checking your calculations for the r_n and f_n values to ensure they are consistent with the given information. If you are still having trouble, it may be helpful to seek guidance from your instructor or a classmate.

Overall, it seems like you have a good understanding of the concept and equations involved in finding the couple vector. With some further clarification and attention to detail, I am confident that you will be able to successfully solve this problem. Keep up the good work!
 

Related to Finding the couple vector of this ice boat

1. How do you find the couple vector of an ice boat?

To find the couple vector of an ice boat, you need to first understand the basic principles of vector analysis. The couple vector is a mathematical representation of the turning effect or torque on an object. To find it, you would need to measure the distance between the point of rotation (fulcrum) and the point where the force is applied (lever arm), and multiply it by the magnitude of the applied force. This will give you the value and direction of the couple vector.

2. What factors affect the couple vector of an ice boat?

The couple vector of an ice boat can be affected by several factors, such as the shape and size of the boat, the force or torque applied, the distance between the force and the point of rotation, and the weight and distribution of weight on the boat. The ice and wind conditions can also impact the couple vector, as well as the angle at which the force is applied.

3. Can the couple vector of an ice boat be negative?

Yes, the couple vector of an ice boat can be negative. This means that the direction of the turning effect or torque is in the opposite direction of the applied force. For example, if the force is applied clockwise and the couple vector is calculated to be counterclockwise, then the couple vector would be negative.

4. How is the couple vector of an ice boat used in navigation?

The couple vector of an ice boat is an important factor in navigation as it helps determine the direction and magnitude of the turning effect on the boat. This information can be used to adjust the sails or rudder to steer the boat in the desired direction. It can also be used to calculate the speed and acceleration of the boat.

5. Can the couple vector of an ice boat change over time?

Yes, the couple vector of an ice boat can change over time. This can happen due to changes in external factors such as wind and ice conditions, as well as changes in the boat's weight distribution or the applied force. It is important for the captain or navigator to continuously monitor and adjust the couple vector to ensure safe and efficient navigation.

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