Solving Coplanar Forces: 8kN, 6kN & 3kN at 45°

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In summary, solving coplanar forces at a 45° angle involves using the principles of vector addition to determine the resultant force. By breaking down the given forces of 8kN, 6kN, and 3kN into their horizontal and vertical components, and using trigonometric functions to find the magnitude and direction of each component, the resultant force can be calculated. This method allows for the determination of the net force acting on an object in a specific direction, and is a fundamental concept in solving problems involving multiple forces acting on a single object.
  • #1
JollyJed
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Homework Statement



I Have to work out the following using mathematical and graphical methods

i have to work out the resultant force of

Vertical = 8kN

Horizontal = 6kN

we were given an example which i understood but with that we were given an extra component of 3kN at 45 degrees, so really not sure how to work this one out!

where would i draw my lines etc
 
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  • #2
JollyJed said:

Homework Statement



I Have to work out the following using mathematical and graphical methods

i have to work out the resultant force of

Vertical = 8kN

Horizontal = 6kN

we were given an example which i understood but with that we were given an extra component of 3kN at 45 degrees, so really not sure how to work this one out!

where would i draw my lines etc
Set up a coordinate system and draw a vertical and horizontal vector (force is a vector) coinciding with the y and x-axis respectively. Would you know how to compute the resultant of these two vectors?
 
  • #3
Thats what i have done mate, horizontal is 6kN and vertical is 8kN , but this is where i am stuck as this is my first time doing it

the example i did was with 3 vectors and i joined the end of the 3'rd to the beginning and measured and got my answer if that makes sense

i have 6 of these but if i can understand this first one the others will be fine!

cheers dude
 
  • #4
JollyJed said:
Thats what i have done mate, horizontal is 6kN and vertical is 8kN , but this is where i am stuck as this is my first time doing it
Do you mean you are stuck after this or are you stuck in finding the resultant of these two vectors?
the example i did was with 3 vectors and i joined the end of the 3'rd to the beginning and measured and got my answer if that makes sense
Joined the end of the third vector to the beginning of what?
 
  • #5
CAF123 said:
Do you mean you are stuck after this or are you stuck in finding the resultant of these two vectors?

Joined the end of the third vector to the beginning of what?

the resultant mate.. to do it graphically you have to draw another line and measure it don't you?

the start point mate (4kN vertical, 4kN horizontal then 3kN at 45 degrees from that, then joined the end of that to the very start of the vertical)
 
  • #6
JollyJed said:
the resultant mate.. to do it graphically you have to draw another line and measure it don't you?

the start point mate (4kN vertical, 4kN horizontal then 3kN at 45 degrees from that, then joined the end of that to the very start of the vertical)
Yes, find the resultant of the vertical and horizontal vectors and (vectorially) add this to the 3kN force at 45 degrees. In this case, it is probably easier to do a proper scale drawing and simply measure the magnitude of the resultant and it's angle from +ve x. You could also use various trig techniques (such as sine/cosine rules).
EDIT: did you notice your vectors have changed from having magnitudes 8 and 6 to 4 and 4?
 
  • #7
CAF123 said:
Yes, find the resultant of the vertical and horizontal vectors and (vectorially) add this to the 3kN force at 45 degrees. In this case, it is probably easier to do a proper scale drawing and simply measure the magnitude of the resultant and it's angle from +ve x. You could also use various trig techniques (such as sine/cosine rules).
EDIT: did you notice your vectors have changed from having magnitudes 8 and 6 to 4 and 4?

thats what i have to do mate, i have to work out using a scaled drawing and using trig, and the vectors 4,4,3 was the one i did at college.. the one I am struggling with at the moment is the 8 and 6 :/

i have drawn the lines on graph paper but have no idea how to get the resultant from it..
 
  • #8
JollyJed said:
thats what i have to do mate, i have to work out using a scaled drawing and using trig, and the vectors 4,4,3 was the one i did at college.. the one I am struggling with at the moment is the 8 and 6 :/

i have drawn the lines on graph paper but have no idea how to get the resultant from it..
Have you done vector addition at all?
See for example
http://www.physicsclassroom.com/class/vectors/u3l1b.cfm
 
  • #9
my scenario look like this picture

u3l1b3.gif


but with one vector 6kN and one 8kN , would my resultant be the red line?
 
  • #10
JollyJed said:
my scenario look like this picture

u3l1b3.gif


but with one vector 6kN and one 8kN , would my resultant be the red line?
Yes.
 
  • #11
CAF123 said:
Yes.

thanks mate :) that's what i thought but didn't seem right..

what do i do after 6(2)+8(2) = 100?

do i square root it?
 
  • #12
JollyJed said:
thanks mate :) that's what i thought but didn't seem right..

what do i do after 6(2)+8(2) = 100?

do i square root it?

Yes, but do you know why?
 
  • #13
CAF123 said:
Yes, but do you know why?

i don't mate.. care to enlighten me?
 
  • #14
JollyJed said:
i don't mate.. care to enlighten me?
You have two vectors that are perpendicular and when you add them together you get a vector going from tail of one to the head of another. You can now think of this as a vector triangle. If the resultant you think of as the hypotenuse of a right angled triangle, you would use Pythagoras to find the length.
Remember your intial problem is a special case (the vectors being orthogonal). You can't always use Pythagoras to find the length.
 
  • #15
CAF123 said:
You have two vectors that are perpendicular and when you add them together you get a vector going from tail of one to the head of another. You can now think of this as a vector triangle. If the resultant you think of as the hypotenuse of a right angled triangle, you would use Pythagoras to find the length.
Remember your intial problem is a special case (the vectors being orthogonal). You can't always use Pythagoras to find the length.

cheers mate :)

how would i work out this one

its a horizontal line with a 45 degree vector going left and a 45 degree vector going right?
 
  • #16
I can't make much sense of that. Can you post a picture?
 
  • #17
CAF123 said:
I can't make much sense of that. Can you post a picture?

number 4 mate

IMAG0273_zps7ac069e0.jpg
 
  • #18
At this point you have to provide an attempt (forum rules) before I can help. What have you learned in the previous problem that can be applied here?
 
  • #19
Well i would draw the first 12kN 45 degree vector, then on top of that i will do the other one, and draw a line from the start to the tip of the 2nd vector - then i would measure that line..

to do it mathematically I am not sure.. as i can't use trig? 12(2)+12(2) and square it?
 
Last edited:
  • #20
JollyJed said:
Well i would draw the first 12kN 45 degree vector, then on top of that i will do the other one, and draw a line from the start to the tip of the 2nd vector - then i would measure that line..

What does that line look like? (i.e is it vertical, horizontal or inclined at some angle)
to do it mathematically I am not sure.. as i can't use trig?

Are the vectors perpendicular in this example?
 
  • #21
CAF123 said:
What does that line look like? (i.e is it vertical, horizontal or inclined at some angle)


Are the vectors perpendicular in this example?

inclined at an angle, so looks like half a kite like

\
/

and no as there's 45 degree angles and not a 90
 
  • #22
Also, in your last example, since force is a vector you necessarily have to specify a direction associated with it. Did you get this? You can do this most easily with a protractor.
 
  • #23
yeah so both lines need to be at a 45degree angle?
 
  • #24
JollyJed said:
inclined at an angle, so looks like half a kite like
That is the addition of the two vectors. Now what is the resultant?

and no as there's 45 degree angles and not a 90

Are you sure they are not 90 degrees with respect to each other? How could you check?

P.s I need to go in a minute or two.
 
  • #25
CAF123 said:
That is the addition of the two vectors. Now what is the resultant?
Are you sure they are not 90 degrees with respect to each other? How could you check?

P.s I need to go in a minute or two.

would the resultant not be from the top of the vector to the start ppoint? so like a straight line down?

well 45 and 45 is 90.. so does that make them perpendicular?

and don't worry mate, you have been a great help :)
 
  • #26
JollyJed said:
would the resultant not be from the top of the vector to the start ppoint? so like a straight line down?
Nearly, but note the direction of the vectors. You want a vector from a tail of one to the head of another.
well 45 and 45 is 90.. so does that make them perpendicular?

and don't worry mate, you have been a great help :)

Superimpose an xy coordinate frame with the leftmost vector in the third quadrant and the right most vector in the fourth quadrant and they both point to the origin. It should be clear now why they are perpendicular. So you can use Pythagoras to find the length and the direction is easy as well.
 
  • #27
could someone help me with 6 please?
 
  • #28
JollyJed said:
could someone help me with 6 please?
5kN + (-5KN) = 0

Remember to add vectors you add head to tail. You're left with just the vertical force.
 
  • #29
FeynmanIsCool said:
5kN + (-5KN) = 0

Remember to add vectors you add head to tail. Your left with just the vertical force.

but there's also 16kn mate?
 
  • #30
JollyJed said:
but there's also 16kn mate?

Correct, so that's the force your left over with. Nothing else can be done.
 
  • #31
FeynmanIsCool said:
Correct, so that's the force your left over with. Nothing else can be done.

and why is 5kn + 5kn 0? it should be 10 surely?
 
  • #32
JollyJed said:
and why is 5kn + 5kn 0? it should be 10 surely?

Well..think of it like this:
If you push against your friend with 70N and he pushes back with 70N, then neither of you will move, so there will be no movement along your X axis (Your force vectors into each other cancel out), but then I can come along perpendicular to you guys (on the Y axis) and push you over with 30N of force, and no force will be back at me (in a very simplified situation, neglecting all other forces) So I'll have a east time doing it. Does that make sense? So the only vector which causes movement in this case would be MY 30N along the Y axis.
 
  • #33
FeynmanIsCool said:
Well..think of it like this:
If you push against your friend with 70N and he pushes back with 70N, then neither of you will move, so there will be no movement along your X axis (Your force vectors into each other cancel out), but then I can come along perpendicular to you guys (on the Y axis) and push you over with 30N of force, and no force will be back at me (in a very simplified situation, neglecting all other forces) So I'll have a east time doing it. Does that make sense? So the only vector which causes movement in this case would be MY 30N along the Y axis.

Yeah i understand mate, but how would you do the working out using the 5,5 and 16kN?
 
  • #34
JollyJed said:
Yeah i understand mate, but how would you do the working out using the 5,5 and 16kN?

Consider first the two 5kN forces. Add these head to tail. What do you get?
You don't have to consider those two forces first, it just simplifies things.
 
  • #35
CAF123 said:
Consider first the two 5kN forces. Add these head to tail. What do you get?
You don't have to consider those two forces first, it just simplifies things.

0kN?
 

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