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Homework Help: Working out the Horizontal and Vertical Components

  1. Sep 26, 2009 #1
    1. The problem statement, all variables and given/known data


    That's an image I have made of the diagram we are given. T1 and T2 are tension 1 and [/i]tension 2[/i] respectively. I need to work out the length of those by:
    a). Maths
    b). Scale Drawing

    2. Relevant equations

    3. The attempt at a solution

    I've tried using sine and cosine, but to no avail =\ (although I know we do have to use them to work it out).

    Please show me how to work it out, but do not give me the answer!.
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Sep 26, 2009 #2
    don't worry, we wouldn't give you the answer even if you asked =P

    the scale drawing you can worry about after doing the maths.

    the key thing here is highlighted in your title, the horizontal and vertical components.

    say you split both t1 and t2 up into horizontal and vertical, you know that because the mass isn't dropping, the upward force (or vertical components) of the tensions must be equal to the weight of the mass. do you see why?
    you also know that it's not moving left or right (or i presume that's given to you), so you know the horizontal components of t1 and t2 must be equal, do you see why? make sure you're happy with these ideas before moving on.

    now, think about what cosine and sin are, have you heard of SohCahToa ? the force itself can be thought of as the hypotenuse of a triangle, with the opposite and adjacent sides as the vertical and horizontal components. having said this, go through every force trying to split it up.

    this is a key idea for ALL of mechanics, you need to nail it
  4. Sep 26, 2009 #3
    First, write out the force equations. This system is in equilibrium, so you need to find:

    [tex]F_x = 0 = ?[/tex]
    [tex]F_y = 0 = ?[/tex]

    Can you see that both tension forces will cancel each other in the x direction? If they are not, then the system is not in equilibrium in the x direction and is in motion, which contradicts the situation. Doing some simple math will tell you that the magnitude of [tex]T_1[/tex] is equal to the magnitude of [tex]T_2[/tex]. Since the angle is the same, the components must be equal in magnitude as well. Since the two tension forces oppose each other in the x direction, they must cancel. We've just proved that the two tension forces cancel twice over!

    Can you see what components of each force contribute to the y direction? Obviously the gravitational force vector is simply itself, but what about the tension? How do you find the component that contributes to the y direction? Remember that the magnitude of [tex]T_1[/tex] = magnitude of [tex]T_2[/tex]? Looking at the y components, do they oppose each other (and therefore cancel), or do they complement each other (and therefore are equal)? From these answers you can build your [tex]F_y = 0 = ?[/tex] equation. Solve for the y-component of T.

    Once you've found some component of the tension force, you can easily find the magnitude. Suppose you had [tex]T\cos(\theta) = a[/tex], where a is any value. You are given [tex]\theta[/tex]. Can you find T?
  5. Sep 26, 2009 #4
    Sorry, if this takes a while for me to understand, but I want to make sure I know this 100%.

    Starting from the top, the title on the worksheet is "Equilibrium" therefore I am pretty sure that it is in equilibrium. Can you explain a bit more on why T1 = T2?
  6. Sep 26, 2009 #5
    well consider it like this,

    What does equilibrium mean? Try to define it. (or look up a definition)

    next, consider what would happen if t2 is BIGGER than t1 and you'll see why you must conclude that they're equal. (do this mathematically if needs be)

    ask as many questions as you need, that's what this forum is here for
  7. Sep 26, 2009 #6
    According to wikipedia: Equilibrium is the condition of a system in which competing influences are balanced

    I understand that completely.

    For my diagram to be in equilibrium, there needs to be a force of 20N above the horizontal line, correct? If so, does T1 + T2 have to add up to 20N?
  8. Sep 26, 2009 #7
    Do you know that [tex]cos(\theta)[/tex] = adjacent / hypotenuse, in a right triangle? If not, now you do. =) Like the other poster said, these trig relationships are key to anything having to do with vectors, which is everything, since vectors are really just motion in non-1-dimensional space, or the entire world.

    Consider the first tension force vector. Draw a line from the tip of the vector to the x axis perpendicular to the axis. Do you see the triangle?

    I don't know if you know this so I'll go over it anyways. A triangle can be represented as three vectors. The two non-diagonal sides can be seen as vectors originating from the intersection. I assume you know how to add vectors. What happens if you add these two vectors? You get the hypotenuse vector! What is this hypotenuse vector in your situation? Here's a picture of a vector-based triangle:


    Your answer to the above question should've been "tension force". Did you get that? If not, do you see why now?

    Do you know the component representation of a vector? It's the x-axis component + the y-axis component. So, by the figure above, we can represent the tension by the horizontal vector and the vertical vector. These horizontal and vertical vectors are the contribution the tension makes in that direction. Now we're getting somewhere - we have a way to find the contribution of the tension along the x-axis! But how do we find it?

    Recall that [tex]cos(\theta) = \frac{A}{H}[/tex], or cos = adjacent over hypotenuse. (This is the SohCahToa thing mentioned - I use Some (Sin) Old Hippie (Opposite/Hypo) Caught (Cos) Another Hippie (Adj/Hypo) Tripping (Tan) On Acid (Opp/Adj))

    Assume we know the magnitude of the tension. Solve for the adjacent, or the horizontal component, or the contribution the tension makes in the x-direction.

    Now, do the triangle thing on the other tension too. You should now have two expressions for the x-component of the tension. You should have the expressions below. If you do not, check your work.

    [tex]T_1x = cos(\theta)T_1[/tex]
    [tex]T_2x = cos(\theta)T_2[/tex]

    Consider the entire system now. Contributions in the y direction should not affect anything in the x direction. If I pull up a string attached to a ball, the ball will not magically start rolling to the left. It will go up, and up only. By your drawing, do you see that the only two forces acting in the x-direction are [tex]T_1x[/tex] and [tex]T_2x[/tex]? Can you also see that they are in opposite directions? If not, stop and think.

    To prove that [tex]T_1 = T_2[/tex], let's consider the opposite case; that is, they aren't equal. Then you can see that the x component of one of [tex]T_1[/tex] and [tex]T_2[/tex] must be greater than the other, since the angle is the same in your drawing. This means the system has a net force in the x-direction somewhere, which by F=ma, puts it into motion. This contradicts the situation (it's in equilibrium), and we have proven that [tex]T_1 = T_2[/tex].

    Sorry, I know I added a lot of background, but I wanted to make sure you understood everything that's going on. It really is an integral concept in kinematics, and if you don't get it now, you'll be in trouble later on.

    In response to your latest question, what part of the triangle do you need to consider now? Remember that if I pull left on the marble, it will not, unless acted upon by Merlin, move up or down.
  9. Sep 26, 2009 #8
    Ok, I now understand why T1 has to be equal to T2 (but I wouldn't know how to work this out if the angles were different).

    I know about vectors, adding vectors, and SOHCAHTOA.

    Before I try out what you have asked (about T1x and T2x) I want to make something else clear.

    T1x + T1y + T2x + T2y = 20

    Would that be correct?
  10. Sep 26, 2009 #9
    Remember the marble analogy? If I pull up, it only moves up. If I pull left or right, it only moves left or right. The x and y components are like my motions of pulling left/right and pulling up/down, respectively. The force of gravity is pulling exclusively down on the object. Do you see what is wrong with your equation?

    Answer: You must consider the net force acting in the x and y directions separately, because they can be thought of as separate. The net force in the x direction is the sum of the vectors pointing exclusively in the x direction. Same for the y direction. Your equation considers the x and y components all in one equation. Remember that equations model real life situations. So, you are modeling the x components and y components influencing each other, which is not true.

    Next question, assuming knowledge of that correct answer. Let's leave the 20N in there. What components of the tension must equal 20N?

    Answer: The y components, because 20N acts in the y direction.

    What is the force equation that describes this? ([tex]F_y = ?[/tex])

    Answer: [tex]F_y = T_1y + T_2y - F_g = 0[/tex]
  11. Sep 26, 2009 #10
    Ahh, and because T1 is equal to T2, and the angles which meet the origin are both 30 degrees, T1x and T1y must be equal length. As T2x is in the negative x direction, and T1x is in the positive x direction, they cancel each other out, only leaving the y components.

    So, T2y + T1y must be equal to 20N, but in an upwards direction, as that would then cancel out with the 20N force downwards therefore reaching equilibrium.

    Is any of that correct?
  12. Sep 26, 2009 #11
    Mostly. "Ahh, and because T1 is equal to T2, and the angles which meet the origin are both 30 degrees, T1x and T1y must be equal length. " is the only gripe I have there. That should say "T1x and T2x must be equal length".

    Everything else is correct though.
  13. Sep 26, 2009 #12
    Sorry, that was actually supposed to say T1x and T2X must be equal length. Is that right?

    What is T2s, Fy and Fg?
  14. Sep 26, 2009 #13
    Remember that, in all 2D equilibria without rotation, the following rules hold:

    Note: measure all angles as standard angles, going CCW from the +x axis.

    The sum of all the x force components is zero.

    The sum of all the y force components is zero.

    The x component of a force A is given by A cos (thetaA) x.

    The y component of a force A is given by A sin (thetaA) y.
  15. Sep 26, 2009 #14
    Fy means the total net force in the y direction (either up or down). Ts was a typo I just fixed xD. Fg is the force due to gravity (20N).
  16. Sep 26, 2009 #15
    Ok, from what I now know:


    And T1 = T2.

    Before I carry on, just want to make sure this is correct. As T1y is equal to T2y, they must each be half of Fg - "10".
    Last edited by a moderator: Apr 24, 2017
  17. Sep 26, 2009 #16
    perfect, their VERTICAL components are 10N each, you're right, now try to work out the horizontal components or just use sin to work out t1
  18. Sep 26, 2009 #17

    sin30 = 10/T2
    T2sin30 = 10
    T2 = 10/sin30
    T2 = 20

    Or have I rearranged it incorrectly?
  19. Sep 26, 2009 #18
    That is correct, to my knowledge.
  20. Sep 26, 2009 #19
    Thanks, so using maths I have worked out that T1 and T2 are both equal to 20N. So, now I need to tackle part b. Am I right in thinking I'll be drawing the sides of the triangles out as vectors, hoping for them to make a closed polygon?
  21. Sep 26, 2009 #20
    I don't know how your teacher wants you to do it using pictures. I would probably draw a downward arrow that's 20cm long, then draw two up arrows that are both 10cm long somewhere else (labeling them as the T_y's), then use a protractor to measure an angle of 30 degrees and form the hypotenuse with another arrow (and fill the bottom x arrow as well, why not), then measure the two hypotenuses (hypotenusii?) and record their length. Should be 20cm.

    Or you could skip right to the triangle building part. Depends on how exact your teacher wants you to be with your drawing.
  22. Sep 26, 2009 #21
    It must be exactly to scale. I've been going through my notes again (after writing up the maths method (twice xD). I've found out that if the system is in equilibrium, drawing the component vectors should assemble a closed polygon. If it does not form a closed polygon, it is not in equilibrium. I should then be able to measure the sides and get an approximate value for each (hopefully somewhat near to my answer of 20 for both T1 and T2 before).
  23. Sep 26, 2009 #22
    Argh, I've just been told by someone that T1 and T2 are not both equal to 20N. What have I done wrong?
  24. Sep 26, 2009 #23
    Everything seems to be in order. Here's how to prove it:

    Consider the force triangle formed by [tex]T_y[/tex] and [tex]T[/tex]. The angle opposite [tex]T_y[/tex] is 30 degrees. The angle formed by [tex]T_y[/tex] and [tex]T_x[/tex] is 90 degrees. The angle formed by [tex]T[/tex] and [tex]T_y[/tex] must be [tex]180-30-90=60[/tex] degrees. This force triangle is otherwise known as the 30-60-90 special right triangle.

    Theory about the 30-60-90 right triangle states that the angle opposite the 30 degree angle can be considered [tex]x[/tex]. The side opposite the 60 degree angle, then, has been found to be [tex]\sqrt{3}x[/tex]. The angle opposite the 90 degree angle, or the hypotenuse, has been found to be [tex]2x[/tex].

    In the case of the tension force triangle, the side opposite the 30 degree angle has been found to be [tex]T_y[/tex]. Thus, [tex]x=10[/tex]. The side opposite the 90 degree angle, or the side in question, is [tex]T[/tex]. Thus, [tex]T=2x=20[/tex]. The final side, or [tex]T_x[/tex], is [tex]10\sqrt{3}[/tex].

    You can also see that the vector [tex]T[/tex] can be represented by [tex]<T_x, T_y>[/tex]. The magnitude of T, or the quantity we are interested in, is given by: [tex]\sqrt{T_x^2 + T_y^2} = \sqrt{100 + 300} = \sqrt{400} = 20.[/tex]

    Simple physical analysis shows that [tex]T_y = 10[/tex] in terms of magnitude. (You have done this part, so I don't need to prove this.)

    This concludes the proof of T = 20. (EDIT: And remember we've proven that [tex]T_1 = T_2[/tex] in terms of magnitude)

    However, if you want signed vectors, consider these statements:
    Let us take the +y direction to be down and the +x to be right. Now consider the direction of [tex]T_1[/tex] and [tex]T_2[/tex]. The y of [tex]T_1[/tex] is facing up, or negative, while the x is left, or negative. Thus, we can say the vector [tex]T_1[/tex] is now [tex]-T_1[/tex], or in component form: [tex]<-Tcos(\theta), -Tsin(\theta)>[/tex], or [tex]<-T_x, -T_y>.[/tex] The magnitude is not changed.

    The y of [tex]T_2[/tex] is facing up, or negative, and the x is facing right, or positive. Thus, we can say the vector [tex]T_2[/tex] is now, in component form: [tex]<Tcos(\theta), -Tsin(\theta)>[/tex], or [tex]<T_x, -T_y>[/tex]. Again, the magnitude is not changed.
  25. Sep 26, 2009 #24
    Ahh that makes more sense now.

    So the maths version is correct, but now I need to work out the drawing, which is harder than I thought. Using a scale of 1CM - 2N, I drew one line straight down 10CM to represent Fg. I then drew on the other 2 lines with 30 degrees interior angles, and well the side lengths weren't close to what I worked out in the maths part. As the triangle is an equilateral, shouldn't all interior angles be 60 degrees? But we're working with 2 30 degrees angles here.

    Sorry for all the questions x_x.
  26. Sep 26, 2009 #25
    I'm not sure about your geometry - could you post a picture? Sorry for the trouble.
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