Understanding Sign Conventions in Tractor and Barge Problem

[cwill53]

Homework Statement

Two tractors are moving at a constant velocity along the shores of a straight canal and pulling a barge by means of two ropes. The tensions in the rope are 160 lbs and 192 lbs. The angle between the ropes is 60 degrees. Determine the water resistance P, opposing the motion of the barge, and the angles A and B which the ropes have to form with the shores, if the barge is required to move parallel to the shores.

Ans. P= 306 lbs; A=33 degrees; B=27 degrees

Relevant Equations

$$\sum \vec{F}=m\vec{a}$$
$$F_{x}=\left \|\vec{F} \right \|cos\phi$$
$$F_{y}=\left \|\vec{F} \right \|sin\phi$$

I've figured out the first part of the question by using components:

However, the second part of the question about finding angles A and B has me stuck.

 

[berkeman]

Yoiks. Your image is sideways and out of focus. Please take a few minutes to type your work into the edit window when you post. Thank you very much.

Also, the next step up is to use the LaTeX tutorial that the PF provides, so that you can post coherent math equations. You can read that simple tutorial under INFO/Help at the top of the page, or under the link at the lower left of the Edit window.

Looking forward to helping you when you reply with something more readable. Thanks.

 

[haruspex]

"The angle between the ropes is 60 degrees. Determine ... the angles A and B which the ropes have to form with the shores,"
Look at what you have wrongly assumed in your diagram.

 

[cwill53]

Yoiks. Your image is sideways and out of focus. Please take a few minutes to type your work into the edit window when you post. Thank you very much.

Also, the next step up is to use the LaTeX tutorial that the PF provides, so that you can post coherent math equations. You can read that simple tutorial under INFO/Help at the top of the page, or under the link at the lower left of the Edit window.

Looking forward to helping you when you reply with something more readable. Thanks.

What I have is:

$\left | T_{1} \right |=160lbs$
$\left | T_{2} \right |=192lbs$
$T_{1x}=-(160lbs)cos60^{^{\circ}}\hat{x}=-80lbs \hat{x}$
$T_{1y}=-(160lbs)sin60^{^{\circ}}\hat{y}=138.56lbs \hat{y}$
$T_{2x}=(192lbs)cos60^{^{\circ}}\hat{x}=96lbs \hat{x}$
$T_{2y}=(192lbs)sin60^{^{\circ}}\hat{y}=166.277lbs \hat{y}$
water resistance $P= T_{1y}+T_{2y}=304.837lbs \hat{y} \sim 306lbs\hat{y}$

 

[cwill53]

"The angle between the ropes is 60 degrees. Determine ... the angles A and B which the ropes have to form with the shores,"
Look at what you have wrongly assumed in your diagram.

Was it that the other angles were also 60 degrees?

 

[jbriggs444]

Was it that the other angles were also 60 degrees?

Yes. They cannot both be 60 degrees.

If they were both 60 degrees then the net cross-river component of the force on the barge would be non-zero.

 

[cwill53]

Yes. They cannot both be 60 degrees.

If they were both 60 degrees then the net cross-river component of the force on the barge would be non-zero.

Thanks a lot, that just told me how to solve the problem.

 

[cwill53]

Yes. They cannot both be 60 degrees.

If they were both 60 degrees then the net cross-river component of the force on the barge would be non-zero.

I actually still don’t understand. After going back to the problem and I’ve still been stuck.

All I’ve done is change the diagram so far.

 

[jbriggs444]

All I’ve done is change the diagram so far.

So you've re-labelled the diagram with $\alpha$ and $\beta$ for the unknown angles. If you try to proceed to the algebra, life is about to become very difficult.

I would like to see you at least set up the equations before I make a suggestion to get out of the difficulty.

 

[cwill53]

So you've re-labelled the diagram with $\alpha$ and $\beta$ for the unknown angles. If you try to proceed to the algebra, life is about to become very difficult.

I would like to see you at least set up the equations before I make a suggestion to get out of the difficulty.

I don't see which equations I would use aside from one about the sum of forces in the x direction, which is
$$\sum F_{x}=-(160lbs)cos\alpha +(192lbs)cos\beta =0$$

Changing the diagram wasn't the only thing I did...I've been at this for a few hours.

 

[jbriggs444]

I don't see which equations I would use aside from one about the sum of forces in the x direction, which is
$$\sum F_{x}=-(160lbs)cos\alpha +(192lbs)cos\beta =0$$

Good.

That is the equation that I hoped and expected you to come up with. The immediate problem is that we now have two variables to solve for: $\alpha$ and $\beta$. We wanted to find $\alpha$. But to do that, we have to somehow get rid of $\beta$.

One approach would be to note that $\alpha$ + 60 degrees + $\beta$ have to add up to 180 degrees. We could solve for $\beta$ in terms of $\alpha$: $\beta=120 - \alpha$ and substitute into your equation, yielding:$$\sum F_x = -160 \text{lbs} \cos \alpha + 192 \text{lbs} \cos (120 - \alpha) = 0$$[I used \text for the lbs and \cos instead of cos to get slightly better fonts. You want variable names to be italicized but text and trig function names not to be. Though your style was very acceptable. Thank you for the TeX]

That is an equation that could, in principle, be solved for $\alpha$. But it looks like a pretty daunting challenge. Maybe fiddling with some trig identities could yield a solution.

But let's try attacking it another way. We know the magnitudes of the two tow-rope forces. We know their relative directions. Surely we can evaluate the magnitude of their vector sum. And we should be able to figure out the direction of that vector sum relative to either of the two rope directions.

Can you attempt that, please?

 

[cwill53]

Good.

That is the equation that I hoped and expected you to come up with. The immediate problem is that we now have two variables to solve for: $\alpha$ and $\beta$. We wanted to find $\alpha$. But to do that, we have to somehow get rid of $\beta$.

One approach would be to note that $\alpha$ + 60 degrees + $\beta$ have to add up to 180 degrees. We could solve for $\beta$ in terms of $\alpha$: $\beta=120 - \alpha$ and substitute into your equation, yielding:$$\sum F_x = -160 \text{lbs} \cos \alpha + 192 \text{lbs} \cos (120 - \alpha) = 0$$[I used \text for the lbs and \cos instead of cos to get slightly better fonts. You want variable names to be italicized but text and trig function names not to be. Though your style was very acceptable. Thank you for the TeX]

That is an equation that could, in principle, be solved for $\alpha$. But it looks like a pretty daunting challenge. Maybe fiddling with some trig identities could yield a solution.

But let's try attacking it another way. We know the magnitudes of the two tow-rope forces. We know their relative directions. Surely we can evaluate the magnitude of their vector sum. And we should be able to figure out the direction of that vector sum relative to either of the two rope directions.

Can you attempt that, please?

Lol no need to be so polite, you're the one assisting me.
What I got for the vector sum, seeing as I can split the angle of 60$^{\circ}$ in half and using that to find the components, was
(16 lbs)$\hat{x}$ + (304 lbs)$\hat{y}$

 

[jbriggs444]

Lol no need to be so polite, you're the one assisting me.
What I got for the vector sum, seeing as I can split the angle of 60$^{\circ}$ in half and using that to find the components, was
(16 lbs)$\hat{x}$ + (304 lbs)$\hat{y}$

Your numbers look correct. But I was after a magnitude and an angle. That will be more convenient for what comes next.

The goal here is to get to a place where we can solve for $\alpha$ or $\beta$.

 

[cwill53]

Your numbers look correct. But I was after a magnitude and an angle. That will be more convenient for what comes next.

The magnitude would be 305.25 lbs and the angle $tan^{-1}(\frac{304.837}{16})$= 87$^{\circ}$.

 

[jbriggs444]

The magnitude would be 305.25 lbs and the angle $tan^{-1}(\frac{304.837}{16})$= 87$^{\circ}$.

That 87 degrees... That is 87 degrees from the x-axis on the assumption that $\alpha$ and $\beta$ are both 60 degrees.

What angle would $\alpha$ need to be so that the resulting angle would be 90 degrees instead.

[It is a very easy question with a very easy answer]

 

[cwill53]

That 87 degrees... That is 87 degrees from the x-axis on the assumption that $\alpha$ and $\beta$ are both 60 degrees.

What angle would $\alpha$ need to be so that the resulting angle would be 90 degrees instead.

[It is a very easy question with a very easy answer]

It would need to be 63 degrees, right?
I think I discovered something I might've done entirely wrong. Is it even possible to use sin(60) and cos(60) as factors for the magnitudes? Shouldn't they be in terms of alpha and beta?

 

[cwill53]

I ended up finding the angles to be -33 degrees and -27 degrees by starting the entire problem over. However, that seems to contradict the fact that the water resistance P= 306 lbs.

 

[jbriggs444]

It would need to be 63 degrees, right?
I think I discovered something I might've done entirely wrong. Is it even possible to use sin(60) and cos(60) as factors for the magnitudes? Shouldn't they be in terms of alpha and beta?

You are correct to be concerned with your method and may have confused yourself.

As I interpreted your working, you implicitly assumed an angle of 60 for $\alpha$ and $\beta$. You then computed the magnitude of the vector sum of the two forces: 87 degrees away from due east. Or three degrees east of north.

That is good as far as it goes. But then the confusion starts. If you want to straighten that thrust out so that it points due north, you have to reduce $\alpha$ by three degrees. Of course, that will mean that you have to increase $\beta$ by three degrees so that the angle between the two remains fixed at 60 degrees.

 

[jbriggs444]

I ended up finding the angles to be -33 degrees and -27 degrees by starting the entire problem over. However, that seems to contradict the fact that the water resistance P= 306 lbs.

Which angles turn out to be -33 and -27 degrees? And what is the meaning of the minus signs?

If the minus signs are popping out of the algebra, that could be a clue to a problem with sign conventions.

 

[kuruman]

Forgive the intervention but, since the shores are parallel to the direction of motion, the angles that the ropes form w.r.t. the shores are the complementary angles to $\alpha$ and $\beta$ shown in the drawing posted in #8. With that redefinition, it is obvious that $\alpha+\beta=60^o$. That plus the cancellation of the transverse components should give numbers for the angles. One can then proceed to find the opposing force.

 

[cwill53]

Which angles turn out to be -33 and -27 degrees? And what is the meaning of the minus signs?

If the minus signs are popping out of the algebra, that could be a clue to a problem with sign conventions.

It must be a matter of signs but it’s odd to say the least. It’s not far off.
I should post my work just for it to be here.