# resultant force

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#1
Two forces of magnitude 6.0 N and 8.0 N act at a point P. Both forces act away from
point P and the angle between them is 40°.

Find the resultant force.
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7 years ago
#2
(Original post by Kolasinac138)
Two forces of magnitude 6.0 N and 8.0 N act at a point P. Both forces act away from
point P and the angle between them is 40°.

Find the resultant force.
What don't you understand?
How far have you got with this?
What theory have you done on this topic?

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#3
(Original post by Stonebridge)
What don't you understand?
How far have you got with this?
What theory have you done on this topic?

Of course.

Essentially so far I've been solving vectors at perpendicular to each other or those which have two components between them. This one looks like this

I thought of adding them tip to tail to make a large resultant, but then I'm not sure what to do. I know Resultant = all components of horizontal and vertical of both vectors, however one only has a horizontal component and adding the others doesn't work.
0
7 years ago
#4
(Original post by Kolasinac138)
Of course.

Essentially so far I've been solving vectors at perpendicular to each other or those which have two components between them. This one looks like this

I thought of adding them tip to tail to make a large resultant, but then I'm not sure what to do. I know Resultant = all components of horizontal and vertical of both vectors, however one only has a horizontal component and adding the others doesn't work.
If you do it graphically it's done as you say by placing them end to end. So you put, say, the 8N tail on the 6N head, keeping the directions the same. The resultant is the vector from the tail of the 6N to the head of the 8N.

You can also think of them in terms of their horizontal and vertical components.
Add the components together to get a total horizontal and total vertical.
The resultant is then found from Pythagoras on those two.
The horizontal force in your diagram has no vertical component so the total vertical component is made up of just the vertical component of the other force.

When you say "it doesn't work" what do you mean?
You need to show your working to find out if you are doing it wrong.
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#5
(Original post by Stonebridge)
If you do it graphically it's done as you say by placing them end to end. So you put, say, the 8N tail on the 6N head, keeping the directions the same. The resultant is the vector from the tail of the 6N to the head of the 8N.

You can also think of them in terms of their horizontal and vertical components.
Add the components together to get a total horizontal and total vertical.
The resultant is then found from Pythagoras on those two.
The horizontal force in your diagram has no vertical component so the total vertical component is made up of just the vertical component of the other force.

When you say "it doesn't work" what do you mean?
You need to show your working to find out if you are doing it wrong.
I think my problem is knowing which one is 8 and which one is 6 N.
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7 years ago
#6
(Original post by Kolasinac138)
I think my problem is knowing which one is 8 and which one is 6 N.
It doesn't matter. The question doesn't specify. The result will be the same.
There is nothing in the question that says one is horizontal, so your answer will be that the resultant is X newtons "at an angle of X to one of the forces, it doesn't matter which.
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#7
(Original post by Stonebridge)
It doesn't matter. The question doesn't specify. The result will be the same.
There is nothing in the question that says one is horizontal, so your answer will be that the resultant is X newtons "at an angle of X to one of the forces, it doesn't matter which.
Ok, so if I take the 8 N to be at 40 degrees, I do

sqrt: 8sin40^2 + 8cos40+6^2 = 13.2 N, which is correct

So at what angle to what vector would that be

Also why does doing that with pythagoras give you the resultant if you don't mind?
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7 years ago
#8
(Original post by Kolasinac138)
Ok, so if I take the 8 N to be at 40 degrees, I do

sqrt: 8sin40^2 + 8cos40+6^2 = 13.2 N, which is correct

So at what angle to what vector would that be

Also why does doing that with pythagoras give you the resultant if you don't mind?
You really need a diagram to see what's going on.

When you have the total horizontal and total vertical components, you have two forces at right angles which form 2 sides of a right angled triangle. The resultant is the hypotenuse.
The angle is found from the right angled triangle using trig. That's why a diagram helps.
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#9
(Original post by Stonebridge)
You really need a diagram to see what's going on.

When you have the total horizontal and total vertical components, you have two forces at right angles which form 2 sides of a right angled triangle. The resultant is the hypotenuse.
The angle is found from the right angled triangle using trig. That's why a diagram helps.
So this is the diagram, isn't it?

How do you get pythagoras from that confuses me.
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7 years ago
#10
I thought you were doing it by adding the horizontal and vertical components?
That's when you use pythagoras. Isn't that what you did?
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#11
(Original post by Stonebridge)
I thought you were doing it by adding the horizontal and vertical components?
That's when you use pythagoras. Isn't that what you did?
Ok, so you'd have the 8cos40 and the 6 along the bottom and a vertical 8sin40, and the hypotenuse of that is the resultant?
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7 years ago
#12
(Original post by Kolasinac138)
Ok, so you'd have the 8cos40 and the 6 along the bottom and a vertical 8sin40, and the hypotenuse of that is the resultant?
Yes. Which is what I said in post #8.
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#13
(Original post by Stonebridge)
Yes. Which is what I said in post #8.
Ok, cool.
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