# Trailer being pulled, where to put origin re: torque?

• lizzyb
In summary, the conversation discusses the problem of calculating the vertical component F_y of the force F on a trailer being pulled by a truck with a force F. The conversation includes equations for the forces in the X and Y directions, as well as a discussion about setting up an equation for torque around the center of mass. Ultimately, the correct equation for torque is determined to be \sum \tau = F_y(L - d) - F_x h - nd = 0, which is used to solve the problem. The conversation ends with the individual expressing gratitude for the help and successfully completing their homework.
lizzyb
Hi. This question is mostly done by a figure; we have a loaded truck of weight W being pulled by a truck with a force F. The CM point on the figure ndicates the location of the center of mass of the trailer and its load.

Suppose the trailer accelerates forward (a>0) or backwards (a<0). What is the vertical component F_y of the force F on the trailer?

For the forces in the X and Y I have:
$$\sum F_x = ma$$ (since there is some acceleration)
$$\sum F_y = n + F_y - W = 0$$

Now, I suppose, I'm to write an equation for the torque, but this one isn't that easy - where do i put the origin? thank you.

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It's usually a good idea when you have an unknown force to calculate torque relative to the point of application of that force (maybe other places too, but at least at that point). This problem does not involve any driving force at the wheel, so the only force at the wheel is normal to the ground as shown in the diagram.

Edit: This was the wrong approach for this problem. See the later posts.

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do you mean put the origin at the center of the wheel? or do you mean put the origin right where the F vector (the one pulling the cart) is pointing off into space?

if i put the origin at the center of the wheel, aren't i supposed to go through the center of mass and if so, how can i include the force vector?

if i put the origin where the F vector goes off into space, how can I include the normal force?

thanks!

I mean put the origin where F is acting. You know the direction and the distances from this point for the other forces acting (n and W). This will tell you the relationship between n and W. You already have the equation that relates n, W and F_y.

Edit: This was the wrong approach for this problem. See the later posts.

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OlderDan said:
I mean put the origin where F is acting. You know the direction and the distances from this point for the other forces acting (n and W). This will tell you the relationship between n and W. You already have the equation that relates n, W and F_y.

so it would be:
$$\sum \tau = w(L - d) - n L = 0$$

So,
$$F_y = w - n = w - \frac{w(L - d)}{L} = \frac{L w - w(L - d)}{L} = \frac{w(L - L + d)}{L} = \frac{wd}{L}$$

hmm, the answers to choose from are things like $$F_y = \frac{W}{2L}(d - \frac{ah}{g})$$

do i need to have the 'bar' of which to calculate the torque go through the center of mass? as in, from F, draw diagonally toward the top-left of the cart? or can I just draw it straight across?

how did they put 'a' in the equation for F_y? In all the answer's I can choose from, 'a' (acceleration) is in all of them.hmmm

lizzyb said:
so it would be:
$$\sum \tau = w(L - d) - n L = 0$$

So,
$$F_y = w - n = w - \frac{w(L - d)}{L} = \frac{L w - w(L - d)}{L} = \frac{w(L - L + d)}{L} = \frac{wd}{L}$$

hmm, the answers to choose from are things like $$F_y = \frac{W}{2L}(d - \frac{ah}{g})$$
I took a wrong turn earlier. You need to calculate the torque about the center of mass and set that to zero. The F has two components. The horizontal component is the only thing that will accelerate the trailer, and the distance from the line of that force component to the CM is given. This force is F_x = ma = a*W/g. The y component of the hitch force (the one you are looking for) also has a known distance, and the normal force has a known distance. You can set up an equation that can be solved for N in terms of the other things from this torque. Then use the other equation for N from the sum of the vertical forces being zero. Set those two expression for N equal and solve for F_y

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when i calculate the torque, do i use a horizontal 'beam' thing?

$$\sum \tau = nd ...$$

since F goes up to the right I'm not sure how to include it.

edited:
$$\sum \tau = -nd + F(L - d) = 0$$
??

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lizzyb said:
when i calculate the torque, do i use a horizontal 'beam' thing?

$$\sum \tau = nd ...$$

since F goes up to the right I'm not sure how to include it.

edited:
$$\sum \tau = -nd + F(L - d) = 0$$
??
(L - d) is the correct distance for the vertical component of F. The horizontal component is a perpendicular distance of h from the CM and don't forget the horizontal component is the only force accelerating the trailer.

this?
$$\sum \tau = F_y(L - d) - F_x h - nd = 0$$
$$\sum F_x = ma = a \frac{W}{g}$$
$$\sum F_y = n + F_y - W = 0$$
??

ok great! for some reason my sign was different than the answers provided but mine was close enough that i tried it. thanks a lot Dan! this question has two more parts but hopefully with this equation i should be able to figure it out! :-)

i did it! thanks everyone! i finished my physics homework - 3 hours to spare (and class in 7 hours :-O {but it's chemistry}). you're an answer to my prayers! yay ! i understand it better, too - i now realize that we're to take the torque around a POINT - not a bar. :-)! good night! thanks again!

lizzyb said:
this?
$$\sum \tau = F_y(L - d) - F_x h - nd = 0$$
$$\sum F_x = ma = a \frac{W}{g}$$
$$\sum F_y = n + F_y - W = 0$$
??
I think your sign problem is here. The two torques from F are both counterclockwise.

$$\sum \tau = F_y(L - d) + F_x h - nd = 0$$

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## 1. What is torque and why is it important when pulling a trailer?

Torque is a measure of the force applied to an object to make it rotate about an axis. When pulling a trailer, torque is important because it determines the amount of rotational force applied to the wheels of the towing vehicle, which in turn affects the ability to move the trailer.

## 2. Where should the origin of the torque be when pulling a trailer?

The origin of the torque should be at the point where the towing vehicle and the trailer are connected. This is typically the hitch on the towing vehicle.

## 3. Can torque affect the stability of the trailer being pulled?

Yes, torque can affect the stability of the trailer being pulled. If the torque is not properly distributed, it can cause the trailer to sway or fishtail, making it difficult to control and potentially causing accidents.

## 4. How can I determine the appropriate amount of torque for pulling a trailer?

The appropriate amount of torque for pulling a trailer depends on several factors, including the weight and size of the trailer, the weight and power of the towing vehicle, and the road conditions. It is best to consult the manufacturer's guidelines or seek advice from a professional to determine the appropriate torque for your specific situation.

## 5. Are there any safety precautions I should take when considering torque for pulling a trailer?

Yes, there are several safety precautions to consider when dealing with torque for pulling a trailer. These include ensuring that the hitch and towing vehicle are properly rated for the weight of the trailer, distributing the weight of the trailer evenly, and regularly checking the torque and trailer connections to ensure they are secure. It is also important to drive at safe speeds and adjust your driving to account for the added weight and potential effects of torque on the trailer.

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