Two ropes in a vertical plane.... sin(theta) or cos(theta)?

[Breadsticks]

Homework Statement

:[/B]
Two ropes in a vertical plane exert equal-magnitude forces on a hanging weight but pull with an angle of 86 degrees between them. What pull does each one exert if their resultant pull is 372N directly upward?

Homework Equations

:[/B]
x-component= magnitude*cos(theta)
y-component= magnitude*sin(theta)
Cy=Ay+By

The Attempt at a Solution

:[/B]
Vectors A, B, and resulting vector C are all in the same plane. If they pull equally, we see the angle between A or B and C is 43 degrees. The x-direction forces cancel, so we are left with the y-direction. If we let A=B, we can simplify. If we add vectors A and B together in the x-direction, we get C=Ay+By, and we can equate it with C=372N. So I have 372=2*magnitude*sin(theta). However, the equation the book uses is 372=2*magnitude*cos(theta). Why? Shouldn't it be sin because it's in the y-direction?

 

[PeroK]

Homework Statement

:[/B]
Two ropes in a vertical plane exert equal-magnitude forces on a hanging weight but pull with an angle of 86 degrees between them. What pull does each one exert if their resultant pull is 372N directly upward?

Homework Equations

:[/B]
x-component= magnitude*cos(theta)
y-component= magnitude*sin(theta)
Cy=Ay+By

The Attempt at a Solution

:[/B]
Vectors A, B, and resulting vector C are all in the same plane. If they pull equally, we see the angle between A or B and C is 43 degrees. The x-direction forces cancel, so we are left with the y-direction. If we let A=B, we can simplify. If we add vectors A and B together in the x-direction, we get C=Ay+By, and we can equate it with C=372N. So I have 372=2*magnitude*sin(theta). However, the equation the book uses is 372=2*magnitude*cos(theta). Why? Shouldn't it be sin because it's in the y-direction?

You need a diagram showing where you have put $\theta$ in relation to the ropes.

You're equations above in section 2 are only valid where $\theta$ is the angle with the horizontal.

 

[Breadsticks]

You need a diagram showing where you have put $\theta$ in relation to the ropes.

You're equations above in section 2 are only valid where $\theta$ is the angle with the horizontal.

Ah, so I rotate the x,y axis by 90 degrees and I'm left with cos(theta).

If I wanted to keep the x,y axis in the traditional location, is there an easy way to modify the equation?

 

[PeroK]

Ah, so I rotate the x,y axis by 90 degrees and I'm left with cos(theta).

If I wanted to keep the x,y axis in the traditional location, is there an easy way to modify the equation?

You'd be better off remembering that sin = opposite/hypotenuse and cos = adjacent/hypotentuse. A triangle, in general, could have any orientation, so its sides will not always be aligned with the x and y axes.

 

[Breadsticks]

You'd be better off remembering that sin = opposite/hypotenuse and cos = adjacent/hypotentuse. A triangle, in general, could have any orientation, so its sides will not always be aligned with the x and y axes.

Okay.

Is there a way to make it work using x-component=magnitude*cos(theta) and y-component=magnitude*sin(theta) if I use two different angles (47 and 133)? I had A use an angle of 133 and B use 47. Then since they cancel each other out, I did magnitude(A)cos(theta)= -magnitude(B)cos(theta) and got 2magnitude(A)=magnitude(B). Then I had magnitude(A)sin(47)+magnitude(B)sin(133), subbed using 2magnitude(A)=magnitude(B) and ended up with magnitude(A)=169N. A would then have a magnitude of 338 and B would have a magnitude of 169. Obviously that's not correct, but I can't tell if it was an assumption or it's just not a viable method.

Edit: This is because it assumes the angle is always the angle from the object it's working on and not just from an axis, right?

 

[William White]

You need to draw a picture

You will find the picture consists of two-right triangles, of exact same dimensions. You know length of the base, you know the angles, calculate the hypotenuese.

 

[BvU]

Okay.

Is there a way to make it work using $\vec V_x= |\vec V| \cos\theta\quad {\rm and} \quad \vec V_y= |\vec V| \sin\theta$ if I use two different angles (47 and 133)? $\quad$

yes

I had A use an angle of 133 and B use 47. Then since they cancel each other out, I did $|\vec A| \cos\theta = -|\vec B|\cos\theta$ and got $2 |\vec A| = |\vec B|. \quad$

wrong twice.

$|\vec A| \cos\theta_A = -|\vec B|\cos\theta_B\Rightarrow |\vec A| = |\vec B|$ because $\cos\theta_A = -\cos\theta_B$, and where the 2 comes from I don't know.

Then I had $|\vec A|\sin{47^\circ} +|\vec B|\sin33^\circ$, subbed using 2|A|=|B| and ended up with |A|=169 N. A would then have a magnitude of 338 and B would have a magnitude of 169. Obviously that's not correct, but I can't tell if it was an assumption or it's just not a viable method.

Edit: This is because it assumes the angle is always the angle from the object it's working on and not just from an axis, right?[/QUOTE]

Try again. And: You did make a drawing I hope ? Drawings are very, very useful.

 

[Breadsticks]

|A→|cosθA=−|B→|cosθB⇒|A→|=|B→||\vec A| \cos\theta_A = -|\vec B|\cos\theta_B\Rightarrow |\vec A| = |\vec B| because cosθA=−cosθB \cos\theta_A = -\cos\theta_B, and where the 2 comes from I don't know.

I must've mistyped on my calculator because I did cos(47)/-cos(133) and got 2. Calculator memory is cleared so I'm not sure what I really typed in. So using the correct substitution, that the magnitudes are equal, I got the right answer. Thank you so much everyone, you've been really helpful!

Try again.

We don't need the angle from the combined vector because it's a combination of two vectors that we already know the positions of and not a separate and unrelated vector, right?

 

[BvU]

We don't need the angle from the combined vector because it's a combination of two vectors that we already know the positions of and not a separate and unrelated vector, right?

In the problem statement you are given that the resultant pull is directly upward. So there is the angle of the sum vector. You use this when you write $|\vec A| \cos\theta_A + |\vec B|\cos\theta_B = 0$: the 0 is the horizontal component of the resultant.

 

[Breadsticks]

In the problem statement you are given that the resultant pull is directly upward. So there is the angle of the sum vector. You use this when you write $|\vec A| \cos\theta_A + |\vec B|\cos\theta_B = 0$: the 0 is the horizontal component of the resultant.

Okay, thank you.

One more question, in this problem's case, the expression for the x-component would be magnitude*tan(theta), right?

 

[BvU]

No. The force points upwards, the $\theta$ you have chosen is wrt the X+ axis ("to the right") so it is $\pi/2$. Just like with the others, the x component is $R\cos\theta$, in this case 0.

Try it out with a resultant at e.g. $\pi/4$

 

[BvU]

I must've mistyped on my calculator because I did cos(47)/-cos(133) and got 2. Calculator memory is cleared so I'm not sure what I really typed in. So using the correct substitution, that the magnitudes are equal, I got the right answer.

Coming back to this one. You are supposed to know and use the fact that $\cos \phi = - \cos (\pi - \phi)$. No calculator should be needed - no typos on that thing can mess it up for you either.

In case you do have to resort to using a calculator: never ever trust the result. So estimate the result any which way you can before you start hammering on the thing.

And the most-often incurred mistake is typing degrees when the setting is for radians (and sometimes vice versa)

 

[Breadsticks]

Coming back to this one. You are supposed to know and use the fact that $\cos \phi = - \cos (\pi - \phi)$. No calculator should be needed - no typos on that thing can mess it up for you either.

In case you do have to resort to using a calculator: never ever trust the result. So estimate the result any which way you can before you start hammering on the thing.

And the most-often incurred mistake is typing degrees when the setting is for radians (and sometimes vice versa)

Yes, thank you.

 

[erickzind]

Yes, thank you.

The same question was my homework today. They just got different degrees (°). Thank you for a very nice explanation.