Drawing Relative Velocity/Triangle Diagrams?

In summary: Thanks!In summary, if person C is holding the rope at an angle of 40° from the horizontal, the balloon will not fly away.
  • #1
kmizzle
6
0

Homework Statement


A balloon with person A in it has a force of gravity of 3000N acting on it. The balloon has an upward force of 3800N due to the hot air in it. Two people are keeping the balloon hovering just above their heads by holding onto ropes attached to each side of the balloon. If person B exerts a force of 540N[Left40°Down] and person C can exert a force of 700N, what is the minimum angle, measured from the horizontal, at which person C should hold the rope in order for the balloon to not fly away?


Homework Equations





The Attempt at a Solution


What I did was:
540sin40=347.12N
Knowing that Fnet"y" should be 0 I did 3800=3347.12N+x
x=452N (person C's horizontal force)

Then this is where I don't know how to solve the problem.

I drew out the relative velocity type triangle diagram by drawing the x component first and then the y component. This would've given me Θ=cos-1(452.88/700)=50°
However, the correct answer is 40°. I didn't know why this was for a long time, but I realized I had to draw the y component and then the x component. This would've given me Θ=sin-1(452.88/700)=40°

So my question is, how do you know whether to draw the x component first or the y component? For previous questions I did I got the correct answer by drawing the x component first, so what makes this one different? :S Thanks!
 
Last edited:
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  • #2
Why did you find 540sin(40)? Where did that 40 come from?
 
  • #3
Averki said:
Why did you find 540sin(40)? Where did that 40 come from?

I did this to find the y component of Person B. 540N was the force, 40 was the angle, and it is sin because it is the y component.
 
  • #4
Your question doesn't make any sense to me. It makes no difference in which order you draw the components, as long as you figure them out correctly.
 
  • #5
kmizzle said:
I did this to find the y component of Person B. 540N was the force, 40 was the angle, and it is sin because it is the y component.

Was the 40 degrees given in the problem statement?
 
  • #6
Averki said:
Was the 40 degrees given in the problem statement?

Yes, it was stated that Person B was exerting a force of 540N[Left40°Down].
 
  • #7
Averki said:
Was the 40 degrees given in the problem statement?

It's right there in the original post!

kmizzle said:

Homework Statement


A balloon with person A in it has a force of gravity of 3000N acting on it. The balloon has an upward force of 3800N due to the hot air in it. Two people are keeping the balloon hovering just above their heads by holding onto ropes attached to each side of the balloon. If person B exerts a force of 540N[Left40°Down] and person C can exert a force of 700N, what is the minimum angle, measured from the horizontal, at which person C should hold the rope in order for the balloon to not fly away?

kmizzle,

I suspect what's tripping you up is WHEN to use sine and when to use cosine, for x and y components. You have to be careful, because it depends on whether your angle is measured from the horizontal or from the vertical. If the angle is measured from the vertical, then the y-component is given by cosine, and the x-component is given by sine. If the angle is measured from the horizontal, then the y-component is given by sine, and the x-component is given by cosine.

If you always just draw the triangle properly and remember the actual defintion of sine and cosine (sine = opposite side/hypotenuse, cosine = adjacent side/hypotenuse), you'll never get it wrong.
 
  • #8
To clear things up, basically, how do I know the first one is how I'm supposed to draw the diagram with regards to the question? What part of the question tells me that I have to draw the y component and then the x?

2883o9.jpg
 
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  • #9
cepheid said:
It's right there in the original post!

Haha... I'm using the PF app and it didn't show up for me... I'm sorry for the confusion.
 
  • #10
cepheid said:
It's right there in the original post!



kmizzle,

I suspect what's tripping you up is WHEN to use sine and when to use cosine, for x and y components. You have to be careful, because it depends on whether your angle is measured from the horizontal or from the vertical. If the angle is measured from the vertical, then the y-component is given by cosine, and the x-component is given by sine. If the angle is measured from the horizontal, then the y-component is given by sine, and the x-component is given by cosine.

If you always just draw the triangle properly and remember the actual defintion of sine and cosine (sine = opposite side/hypotenuse, cosine = adjacent side/hypotenuse), you'll never get it wrong.

Ok thanks, I think I have it figured out!
 
  • #11
kmizzle said:
To clear things up, basically, how do I know the first one is how I'm supposed to draw the diagram with regards to the question? What part of the question tells me that I have to draw the y component and then the x?

2883o9.jpg

If you look back at the initial problem statement, it asks for the minimum angle measured from the horizontal, not the vertical. Technically, the second triangle in your image is the correct diagram. However, by using the cosine you are finding the x-component of the 700N force, which is not useful in answering this question. If you follow cepheid's advice, which trigonometric function will give you the y-component of the 700N force?
 
  • #12
Averki said:
If you look back at the initial problem statement, it asks for the minimum angle measured from the horizontal, not the vertical. Technically, the second triangle in your image is the correct diagram. However, by using the cosine you are finding the x-component of the 700N force, which is not useful in answering this question. If you follow cepheid's advice, which trigonometric function will give you the y-component of the 700N force?

Yes I see that I need to use sin now and the reason for it! Thank you so much!
 

1. What is a relative velocity/triangle diagram?

A relative velocity/triangle diagram is a graphical representation used to solve problems involving motion in two dimensions. It shows the velocities and directions of two objects moving relative to each other.

2. How do you construct a relative velocity/triangle diagram?

To construct a relative velocity/triangle diagram, you first need to identify the velocities and directions of the two objects involved. Then, draw a scale diagram with the velocities as vectors, making sure to label the vectors with their magnitude and direction. Finally, use vector addition to find the resulting velocity.

3. What is the purpose of using relative velocity/triangle diagrams?

The purpose of using relative velocity/triangle diagrams is to simplify and solve problems involving motion in two dimensions. It allows you to visually see the velocities and directions of the objects involved and use vector addition to find the resulting velocity.

4. Can relative velocity/triangle diagrams be used for any type of motion?

Yes, relative velocity/triangle diagrams can be used for any type of motion, as long as it involves two objects moving relative to each other. This includes linear, circular, and projectile motion.

5. What are some common mistakes when using relative velocity/triangle diagrams?

Some common mistakes when using relative velocity/triangle diagrams include using incorrect units, not properly labeling the vectors, and not taking into account the direction of the resulting velocity. It is important to carefully follow the steps for constructing the diagram and double check your work to avoid these mistakes.

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