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

[JollyJed]

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?

 

[FeynmanIsCool]

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.

 

[JollyJed]

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?

 

[CAF123]

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.

 

[JollyJed]

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?

 

[CAF123]

0kN?

Yes. Now what are you left with?

 

[JollyJed]

like that?

and left with the 16kN mate?

 

[CAF123]

No. Forces are vectors and so you have to put arrows on those lines to give a direction. Taking the two 5kN forces, we have already said that the resultant of those two is 0 (or the zero vector to be more accurate). But you still have a 16kN force acting straight down. So what do you get in the end?

 

[JollyJed]

No to the picture or no to the 16kN left over?

what i was told at college was the vectors could go in any direction, and would always give the same result..

 

[CAF123]

No to the picture or no to the 16kN left over?

The picture. You are right in that 16kN is all that is left.

what i was told at college was the vectors could go in any direction, and would always give the same result..

If I understand you correctly, that is simply not true. If the 5kN forces were pointing in the same direction, their resultant would not be zero. Hence direction matters.

 

[JollyJed]

The picture. You are right in that 16kN is all that is left.

If I understand you correctly, that is simply not true. If the 5kN forces were pointing in the same direction, their resultant would not be zero. Hence direction matters.

Ahh yes i understand what you mean now mate! so if one of the 5kN forces was facing the opposite way to the other i would be right?

so when drawing it graphically would i draw it the same? and obviously 'measure' the 16kn line?

cheers for all the help mate!

 

[CAF123]

What I think you mean is that if you have a vector pointing in some direction, you can translate that vector anywhere in space, provided you maintain it's orientation. E.g a vector with it's tail at the origin and it's head at 2 units pointing along positive x axis. Clearly, that has a length of 2 units. Now exactly the same vector is given by a vector that has it's tail at, say, 5 units along +ve x from origin and it's head 7 units along +ve x from origin. The length is still 2 and the direction is unchanged. This is probably what you meant.

 

[CAF123]

Ahh yes i understand what you mean now mate! so if one of the 5kN forces was facing the opposite way to the other i would be right?

No, because you added them head to head.

so when drawing it graphically would i draw it the same? and obviously 'measure' the 16kn line?

cheers for all the help mate!

The resultant is simply a vertical line pointing in the -ve y direction with magnitude 16kN.

 

[JollyJed]

that was the example i was given (at the bottom)

so why can't i use that here?

 

[CAF123]

Why can't you use what here?

 

[JollyJed]

Why can't you use what here?

Move the vectors like a did in the previous pictures? I am pretty sure the arrows are just pointing to where the origin is, which they were in the example but i could still move them..

 

[CAF123]

Move the vectors like a did in the previous pictures? I am pretty sure the arrows are just pointing to where the origin is, which they were in the example but i could still move them..

I don't see any arrows on your picture, but provided you always add vectors head to tail, you will get the right answer.

Looking at the solution at the top of your photo, it looks like one 4KN force is pointing along +ve y, the other along +ve x. When you add these head to tail, you get that diagonal of length √{4)²+(4)² at 45 degrees wrt positive x. Then this is added head to tail to the 3kN force.

 

[JollyJed]

Yes which is how i worked out the answer for that, so how do you do 6? from what you have said i don't change it at all and 16kN will be the resultant.. but how do i do it mathmatically?

5kN + 5kN = 0kN meaning 16kN is the resultant?

 

[FeynmanIsCool]

16kN is the resultant?

Yes, 16kN is the Resultant vector magnitude.

 

[JollyJed]

I cracked it cheers lads, you make 3 lines of a rectangle with the 3 you are given and the resultant is the last line :) 16kn :D