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Magnitude & Direction of Resultant Force & Equilibrant Force

  • #1
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Homework Statement


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.

Homework 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:

Answers and Replies

  • #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.
 
  • #3
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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.
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.
 
  • #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?
 
  • #5
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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?
Updated with images.
 
  • #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 = ~?##
 
  • #7
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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.
 
  • #8
gneill
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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.
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.
 
  • #9
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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.
 
  • #10
gneill
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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.
Well, we're here to help when and where we can. And yes, I admit that I did note the irony :smile:
So I'm assuming that they should look something like this?

F1x = 5kN
F1y = 8.66kN

F2x = 10.6kN
F2y = 10.6kN
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:
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.
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?
 
  • #11
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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?
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:
 
  • #12
gneill
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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
That looks better.
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 beleive I also get confused at this point :sorry:
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.
 
  • #13
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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:
 
  • #14
haruspex
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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:
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?
 
  • #15
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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
 
  • #16
gneill
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Okay, so you now have the resultant force. What's your next step?
 
  • #17
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Okay, so you now have the resultant force. What's your next step?
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.
 
  • #18
gneill
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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.
Almost. Check the angle: What are the signs of the force components? Did you include them in the determination of the angle?
 
  • #19
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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?
Oops, forgot the minus.

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

Becoming, 10.8kN and 301°?
 
  • #20
gneill
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Oops, forgot the minus.

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

Becoming, 10.8kN and 301°?
-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?
 
  • #21
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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?
Ah okay thanks, and I suppose I now need to find the magnitude and direction of the equilibrant force. I know that for an equilibrium to be made, it needs to have equal forces acting upon it but I'm not sure what steps to take, would the force just be 10.8kN and the angle 59° instead of -59°?
 
  • #22
gneill
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To counter the resultant force the equilibrant force must have the opposite direction to the resultant force. Just negating the angle won't do (think, for example, about small angles close to the x-axis). The angles should end being 180° apart...
 
  • #23
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To counter the resultant force the equilibrant force must have the opposite direction to the resultant force. Just negating the angle won't do (think, for example, about small angles close to the x-axis). The angles should end being 180° apart...
Yeah, not sure why I decided to whack a minus in front of it and assume it 180° hahaha. Would it be 118° instead?
 
  • #24
gneill
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Yeah, not sure why I decided to whack a minus in front of it and assume it 180° hahaha. Would it be 118° instead?
-59 + 180 = ?
 
  • #25
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-59 + 180 = ?
121°.. I most definitely misclicked.. I added the two numbers together.
 

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