A Horse Pulls a Barge in a River; What is the force of the water on the Barge?

[emmy]

Homework Statement

In the figure, a horse pulls a barge along a canal by means of a rope. The force on the barge from the rope has a magnitude of 7830 N and is at the angle θ = 18° from the barge's motion, which is in the positive direction of an x-axis extending along the canal. The mass of the barge is 9500 kg, and the magnitude of its acceleration is 0.12 m/s2. What are (a) the magnitude and (b) the direction (measured from the positive direction of the x axis) of the force on the barge from the water?
http://edugen.wiley.com/edugen/courses/crs4957/art/qb/qu/c05/q48.jpg

2. The attempt at a solution

[PLAIN]http://i629.photobucket.com/albums/uu15/amorxamor/5-42.jpg

So (a) the magnitude is 6690 N
(b) the direction is 18 degrees from +x direction (upstream)

I'm at a loss. I think this is how you do it...?

 

[gneill]

If the barge's motion is strictly in the x-direction as stated in the problem, then there should be no acceleration in the y-direction. That means that the water is providing a force to counter the rope's y-component of force. This is easily calculated from the given rope tension and direction.

The net acceleration is in the x-direction, so the rope's x-direction force, countered by the water's x-direction force, must yield the desired acceleration (f_net/m).

By the way, I like your diagram. It has the look of a whiteboard sketch. What did you use to produce it?

 

[emmy]

If the barge's motion is strictly in the x-direction as stated in the problem, then there should be no acceleration in the y-direction. That means that the water is providing a force to counter the rope's y-component of force. This is easily calculated from the given rope tension and direction.

The net acceleration is in the x-direction, so the rope's x-direction force, countered by the water's x-direction force, must yield the desired acceleration (f_net/m).

By the way, I like your diagram. It has the look of a whiteboard sketch. What did you use to produce it?

That's what I was thinking that the water had to be countering the force in the y direction, since it's definitely traveling in a straight line:

net ay=0, water ay= -horse ay = 7830N/9500kg =0.824m/s^2
Fy=m.ay.sin(θ)= 7830sin(18)= 2419.6N
net ax=0.12m/s^2
Fx= m.ax.cos(θ)= (9500kg)(0.12m/s^2)cos(18)= 1084.2 N

from these values, the magnitude is:
sqrt(2419.6^2 + 1084.2^2)= 2651.4 N

and then θ:
tan^-1(1084.2/2419.6)= 22.2 degrees with respect to +x?

I must've gotten something wrong somewhere; shouldn't the water be traveling in some direction close to opposite the barge, especially since it's exerting a y component of force away from the horse...


and thanks! I actually made it on photoshop with a tablet :shy:

 

[S_Happens]

That's not the only solution for theta. There is indeed another solution that might seem a lot more plausible.

 

[emmy]

That's not the only solution for theta. There is indeed another solution that might seem a lot more plausible.

Hmm.. I know that I could technically do the arccos or arcsin functions, but since I have the x and y components... it won't give the correct angle?

Either way I need both the x and y components of the force of the water ):

 

[gneill]

That's what I was thinking that the water had to be countering the force in the y direction, since it's definitely traveling in a straight line:

net ay=0, water ay= -horse ay = 7830N/9500kg =0.824m/s^2
Fy=m.ay.sin(θ)= 7830sin(18)= 2419.6N
net ax=0.12m/s^2
Fx= m.ax.cos(θ)= (9500kg)(0.12m/s^2)cos(18)= 1084.2 N

Not sure I follow that last line. The net force in the x-direction can be obtained from the acceleration and the mass of the barge. No cosine required. Where the cosine comes in is to find the contribution to the x-direction force of the rope tension. Then, if Fx is the net force in the x-direction (producing the acceleration),

Fx = Fx_rope + Fx_water = m a_x

Fx_water = Fx - Fx_rope

With Fx_water and Fy ( = Fy_water) in hand, you have the components of the force due to the water.

I must've gotten something wrong somewhere; shouldn't the water be traveling in some direction close to opposite the barge, especially since it's exerting a y component of force away from the horse...

I suspect that the barge is employing a rudder...

 

[emmy]

Not sure I follow that last line. The net force in the x-direction can be obtained from the acceleration and the mass of the barge. No cosine required. Where the cosine comes in is to find the contribution to the x-direction force of the rope tension. Then, if Fx is the net force in the x-direction (producing the acceleration),

Fx = Fx_rope + Fx_water = m a_x

Fx_water = Fx - Fx_rope

With Fx_water and Fy ( = Fy_water) in hand, you have the components of the force due to the water.

I suspect that the barge is employing a rudder...

Fy=m.ay.sin(θ)= 7830sin(18)= 2419.6N

Fx= (9500kg)(0.12m/s²)= 1140 N
Fx_rope= 7830cosθ= 7830cos(18)= 7446.8N
Fx_water= unknown, this is what we're solving for...

Fx_water= Fx - Fx_rope= (1140-7446.8)N= -6306.8 N (which makes sense since the water is flowing in the -x direction)

Then the angle θ would be
arctan (Fy/Fx)= θ
arctan(2419.6/-6306.8)= -21^o
θ= -21.0^o

 

[gneill]

The fy_water should be negative too. It's countering the rope's y-component, which is acting in the +y direction. That puts the water force direction in the 3rd quadrant, with an angle in the neighborhood of -160 degrees (you can calculate a precise value).

 

[emmy]

The fy_water should be negative too. It's countering the rope's y-component, which is acting in the +y direction. That puts the water force direction in the 3rd quadrant, with an angle in the neighborhood of -160 degrees (you can calculate a precise value).

oh but I realized that the question asks for the magnitude of the force on the barge from the water, so that would be

(2419.6²+6306.8²)^1/2 = 6755.0 N correct?


and then for the angle, did you just subtract 18^o-180^o to get the water to be flowing 180^o away from the Tension in the rope to the horse? It would make sense that the water is flowing that way because the barge is moving in a straight line...

 

[gneill]

The force magnitude looks good (at least, it's the same as what I arrived at).

For the angle of the force due to the water, I used the individual components in atan() and sorted out the quadrant from the signs of the components. It turned out to be not quite 180° from the rope angle.

 

[emmy]

The force magnitude looks good (at least, it's the same as what I arrived at).

For the angle of the force due to the water, I used the individual components in atan() and sorted out the quadrant from the signs of the components. It turned out to be not quite 180° from the rope angle.

Great!

So i ended up with about 201^o from the +x direction... from doing arctan(2419.6/6306.8) and then adding 180^o to the calculator answer since it should be in the III Quadrant.

 

[gneill]

Looks good!

 

[emmy]

Looks good!

Hmm, well the magnitude was correct, but the angle was wrong when I submitted it

 

[gneill]

Hmm, well the magnitude was correct, but the angle was wrong when I submitted it

That's disconcerting. Perhaps they were looking for the angle measured in the other direction? -159.0° ? If so, the software is deficient.

 

[emmy]

There have been complaints about the software before, maybe they did want -159... it was a logical answer...

Thank you very much for all your help!

 

[gneill]

My pleasure.