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Magnitude & Direction of Resultant Force & Equilibrant Force

  1. Nov 18, 2016 #1
    1. The problem statement, all variables and given/known data
    I need help as to how to find the magnitude and direction of the resultant force, and the equilibrant force.
    qlD4eVM.png
    Any and all help is greatly appreciate as I'm stressing myself out over it and I'm not sure what to do.
    Thanks, Jamie.

    2. Relevant equations
    First question respectively:
    JAsghkI.png
    Second question repsectively:
    r0wCrVF.png

    3. The attempt at a solution

    My calculations thus far..
    For first question:
    BE7rXg4.png

    For second question:
    zBDmcBb.png
    2rO5zP3.png
    http://imgur.com/zBDmcBb [Broken]
    http://imgur.com/2rO5zP3 [Broken]
     
    Last edited by a moderator: May 8, 2017
  2. jcsd
  3. Nov 18, 2016 #2

    gneill

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    You need to show an attempt at solution before help can be offered. What have you tried? What calculations have you done?

    Have you provided the complete problem description? It's not clear from what you've written how an answer is to be specified.
     
  4. Nov 18, 2016 #3
    I updated the first link to show the description which accompanied the diagram, and added another link showing the question related to the diagram. I tried working out the x and y components, and once I done that I didn't know what to do afterwards. I used pythagoras' theorem to try and find the magnitude but that was for one "triangle" I guess but I need to find the magnitude ect of the entire "tool" and to which I have had no luck whatsoever.
     
  5. Nov 18, 2016 #4

    gneill

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    Show your calculations so far. For example, for each force on the diagram show your calculation for its x and y components.

    What does one do with a set of forces in order to find the resultant (or net) force?
     
  6. Nov 18, 2016 #5
    Updated with images.
     
  7. Nov 18, 2016 #6

    gneill

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    Please post your work in new posts rather than updating the original post. Otherwise the conversation can't be followed logically by new readers. Also, it's preferable to type out your work so that helpers can quote portions of it if they spot an error they wish to comment on. Trying to point out issues in an image can be frustrating.

    Some of your images are "broken" (can't be viewed), so I can't tell what they might have contained. But it looks to me as though you haven't identified all the forces. Let's label them on the image:

    upload_2016-11-18_17-22-25.png

    The two men are each applying a force, F1 = 10 kN and F2 = 15 kN. Gravity contributes Fg = 5 kN and the crane is providing an upward force of Fc = 15 kN. So that's four forces to break down into their components before you can sum them to yield your resultant. Make a list and fill out the components:

    ##F1_x = ~?##
    ##F1_y = ~?##

    ##F2_x = ~?##
    ##F2_y = ~?##

    ##Fg_x = ~?##
    ##Fg_y = ~?##

    ##Fc_x = ~?##
    ##Fc_y = ~?##
     
  8. Nov 18, 2016 #7
    I do apologise, first time posting on this forum. Here's my attempt at calculating the components of each of the forces:

    F1x = 4.33kN
    F1y = 4.33kN

    F2x = 7.49kN
    F2y = 7.49kN

    Fgx = 0kN
    Fgy = -5kN

    Fcx = 0kN
    Fcy = 15kN

    Not sure if last 2 forces are correct.
     
  9. Nov 18, 2016 #8

    gneill

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    Okay, the last two forces are correct. But the first two are not!

    Looking back at the images that you added to the first post it looks like you're applying trig twice for some reason. You extracted the components correctly and then took those values and multiplied them by sines and cosines again. That's not right.

    You actually extracted the components of F1 and F2 in this image from your post:
    proxy.php?image=http%3A%2F%2Fi.imgur.com%2FBE7rXg4.png

    You've used arrows instead of signs to indicate their directions, but the numerical values are fine. You can use these values (with appropriate signs) to update the force list.

    Taking the force F1 as an example you can draw the force and its components:
    upload_2016-11-18_19-34-27.png
    and that's it for extracting the components of F1.
     
  10. Nov 18, 2016 #9
    First off, I'd like to say thanks for the help thus far, only decent help I've recieved, and that's ironic that the first 2 are incorrect hahaha.

    So I'm assuming that they should look something like this?

    F1x = 5kN
    F1y = 8.66kN

    F2x = 10.6kN
    F2y = 10.6kN

    Fgx = 0kN
    Fgy = -5kN

    Fcx = 0kN
    Fcy = 15kN

    So I did infact take the first components and then do the same calculations with those answers, oops hahaha. So instead of doing "√4.332+4.332" which I done above, I should do "√52+8.662" and then the same with F2? Sorry if I seem clueless, but I am exactly that, but hey that's why I'm here.
     
  11. Nov 18, 2016 #10

    gneill

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    Well, we're here to help when and where we can. And yes, I admit that I did note the irony :smile:
    The digits are now correct, but you need to set the correct signs on the components. F1 is pointing down and to the left, putting it in the 3rd quadrant of the Cartesian plane. What signs should you give its components? F2 is pointing down and to the right, putting it in the 4th quadrant. What signs will its components have?

    The next two forces are fine, having the right signs on the components:
    Before calculating magnitudes you need to compose the resultant (or net) force. What do you know about finding the net force when several forces are acting?
     
  12. Nov 18, 2016 #11
    I guess we need to add the negative signs in where suitable, since F1 is pointing down and to the left, both axis must be negative, and as for F2, only one axis is negative, in this case it is y so a negative sign must be added.

    F1x = -5kN
    F1y = -8.66kN

    F2x = 10.6kN
    F2y = -10.6kN

    As for resulting force, do we add the acting forces together? I think I used the forumla
    R = F1+F2+... or in our case, I think, R = F1x + F1y +...
    I'm not sure if I add just the two forces e.g F1x and F2y or if I add both them and the 10kN we were originally told about, I believe I also get confused at this point :sorry:
     
  13. Nov 18, 2016 #12

    gneill

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    That looks better.
    The reason for breaking the forces into their components is so that you can add the vectors together. You add vectors by adding up their like-components. Sum up all the x-components to give you the x-component of the resultant force. Similarly, sum up all the y-components to give you the y-component of the resultant force. In the end you have the two components that make up the resultant.
     
  14. Nov 18, 2016 #13
    Would it be like this for the x-components:

    F1x = -5kN
    F2x = 10.6kN
    Fgx = 0kN
    Fcx = 0kN

    Rx = -5 + 10.6
    Rx = 5.6kN

    And for y-components:
    F1y = -8.66kN
    F2y = -10.6kN
    Fgy = -5kN
    Fcy = 15kN

    Ry = -8.66 + -10.66 + -5 + 15
    Ry = -9.32kN

    Leaving us with Rx = 5.6kN and Ry = -9.32kN

    I was convinced I had to do it all seperately, that's why I labeled one part 'x' and another part 'y' on the diagram :rolleyes:
     
  15. Nov 18, 2016 #14

    haruspex

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    Looks right.
    I note that it asks for the magnitude and direction of the equilibrant force. Of course, to achieve equilibrium there must also be no net torque, so the line of application of the equilibrant force is also important. Strange that it does not ask for that. Is there perhaps another part to the question?
     
  16. Nov 19, 2016 #15
    Yes, there were 4 questions in total, I thought I had done the first two but I had only partially done the 2nd one and I was seriously struggling with the 3rd and 4th questions. These are all the questions that came with the diagram:
    OiCxQ8N.png
     
  17. Nov 19, 2016 #16

    gneill

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    Okay, so you now have the resultant force. What's your next step?
     
  18. Nov 19, 2016 #17
    To find the magnitude and direction of the resultant force.

    Magnitude:

    R = √Rx2 + Ry2

    R = √
    5.62 + 9.322 = 10.8kN

    Direction:

    θ = tan-1 Ry /Rx

    θ = tan-1 9.32 / 5.6 = 59°

    I'm near certain those calculations are correct.
     
  19. Nov 19, 2016 #18

    gneill

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    Almost. Check the angle: What are the signs of the force components? Did you include them in the determination of the angle?
     
  20. Nov 19, 2016 #19
    Oops, forgot the minus.

    θ = tan-1 -9.32 / 5.6 = -59 which would be 360 - 59 = 301°?

    Becoming, 10.8kN and 301°?
     
  21. Nov 19, 2016 #20

    gneill

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    -59° is adequate. Usually angles are "normalized" to the range -180° ≤ θ ≤ 180° for presentation.

    So now you have the resultant force in both component form and polar form (magnitude and angle). What's next?
     
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