Vertical & Horizontal Components of Acceleration (Upward Force)

AI Thread Summary
The discussion revolves around a physics problem involving a barge being towed by horses, focusing on the vertical and horizontal components of acceleration. The horses exert a force at an angle, leading to confusion about the direction of acceleration. It is clarified that the barge accelerates horizontally along the canal due to the net force from the horses, while vertical forces like buoyancy and gravity balance each other. The correct approach involves separating the forces into their components and applying Newton's second law to determine the forces acting on the barge. Ultimately, the resolution highlights the importance of understanding the directional components of forces in analyzing motion.
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Homework Statement


In the picture below, we see a scene from the days of yore. This is called a tow path and the situation is that a barge is being towed by some horses along the side of a canal. Let's say the horses pull with 6200 N of force at an angle of θ=16∘. If the barge has a mass of 2528 kg, and magnitude of its acceleration is 1.47 m/s2, find the magnitude of the force on the barge from the water.

Here's a picture of the angle between the rope and the barge. Assuming I'm reading it correctly, it's with respect to the vertical component (so the barge isn't too far behind the horses)

tow-path.jpg


Homework Equations



F sin θ = Fx
F cos θ = Fy
F = ma

The Attempt at a Solution



Just to start off, I want to say that I figured out the answer, but I'm not sure why the answer is what it is.

I tried to calculate the horizontal component of the force applied by the horses (which would be 6,200 * sin 16, or 1,709.95 N). Then, the total force on the barge would be 3,716.16 N (due to F = ma). So the force on the barge should be the difference between the total force and the force applied by the horses, which was 2,006.21 N.

I sent this explanation to my professor, and he said that the acceleration is only happening in the y-direction. But that doesn't make sense, because the horses are dragging the barge in the x-direction. He said Fnet x would be 6200sin16 - Fwater x = 0 (which makes sense if there's no acceleration on the x-axis). Then he told me to figure out the equation for Fnet y, which would be 6200cos16 - Fwater y = 3,716.16 N (because F=ma).

So then I solved for Fwater x and Fwater y, and used the Pythagorean theorem to calculate that the net force of the water was 2820.40 N, which was the correct answer.

So ultimately, my question is: Why is the barge accelerating vertically instead of horizontally? It's not like the horses are pulling it out of the water. Unless the buoyant force is normally canceled out by the gravitational force, but the upward force exerted by the horses is causing upward acceleration. But then why is there no horizontal acceleration?
 
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Careful what you mean by horizontal and vertical. I think these refer to the horizontal and vertical directions in the diagram: horizontal being side-to-side in the canal, vertical being along it. After all, the horses are pulling the barge down the canal, not out of it. There is some force to the side of the canal, but this gets canceled by the water or the side of the canal pushing back, so there is only acceleration down the canal due to the force along the canal.
 
jackarms said:
Careful what you mean by horizontal and vertical. I think these refer to the horizontal and vertical directions in the diagram: horizontal being side-to-side in the canal, vertical being along it. After all, the horses are pulling the barge down the canal, not out of it. There is some force to the side of the canal, but this gets canceled by the water or the side of the canal pushing back, so there is only acceleration down the canal due to the force along the canal.

Ah, that makes sense why the blue strip in the diagram is where it is. Thank you so much for your help. :smile:
 
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