My Math Forum Vector magnitude

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July 15th, 2019, 09:43 AM   #1
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Vector magnitude

Two forces, each equal to P, act at right angles. Their effect may be neutralized by a third force acting along their bisector in the opposite direction with a magnitude of___________
in this diagram, the resultant vector is the opposite direction, So what should I do the addition of vector or subtraction of vector? I am confused.
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 July 15th, 2019, 09:57 AM #2 Senior Member   Joined: Jun 2019 From: USA Posts: 120 Thanks: 40 The effect is neutralized by a third force. That means Pi + Pj + F = 0. Resolve it into the i component and the j component, that will give you the two components of F, then find the magnitude with Pythagoras. Thanks from Indranil
July 15th, 2019, 10:52 AM   #3
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 Originally Posted by DarnItJimImAnEngineer The effect is neutralized by a third force. That means Pi + Pj + F = 0. Resolve it into the i component and the j component, that will give you the two components of F, then find the magnitude with Pythagoras.
then will be the components, -i or +i and -j or +j because the resultant vector is the opposite side? I am confused.

 July 15th, 2019, 11:07 AM #4 Senior Member   Joined: Jun 2019 From: USA Posts: 120 Thanks: 40 OK, let's let (+)x or i be to the right, and (+)y or j be up in the diagram. x components: $\displaystyle P + 0 + F_x = 0 \rightarrow F_x = -P$ y components: $\displaystyle 0 + P + F_y = 0 \rightarrow F_y = -P$ Does that clear it up? You could shift the negative signs around, and say something like the leftward component of F equals the rightward component of P, and the downward component of F equals the upward component of the other P. This would get you to the same answer; you just have to keep track of how you defined things. Personally, I recommend the first way -- defining a positive direction for each axis, and letting the components be positive or negative as needed. This is standard in maths/physics, and I find it much easier to keep things straight this way. Thanks from Indranil
July 15th, 2019, 01:31 PM   #5
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Quote:
 Originally Posted by Indranil the resultant vector is . . .
You need its magnitude.

 July 15th, 2019, 01:44 PM #6 Senior Member   Joined: Jun 2019 From: USA Posts: 120 Thanks: 40 It's been mentioned how to find the magnitude. For beginners, I think the first step should always be to find the individual components, and he may as well practice getting the signs correct, even if they don't matter to the specific question at hand.
July 15th, 2019, 08:57 PM   #7
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Quote:
 Originally Posted by DarnItJimImAnEngineer OK, let's let (+)x or i be to the right, and (+)y or j be up in the diagram. x components: $\displaystyle P + 0 + F_x = 0 \rightarrow F_x = -P$ y components: $\displaystyle 0 + P + F_y = 0 \rightarrow F_y = -P$ Does that clear it up? You could shift the negative signs around, and say something like the leftward component of F equals the rightward component of P, and the downward component of F equals the upward component of the other P. This would get you to the same answer; you just have to keep track of how you defined things. Personally, I recommend the first way -- defining a positive direction for each axis, and letting the components be positive or negative as needed. This is standard in maths/physics, and I find it much easier to keep things straight this way.
No
The components should be Fsin theta and Fcos theta.

 July 15th, 2019, 11:24 PM #8 Senior Member   Joined: Apr 2014 From: UK Posts: 953 Thanks: 340 No, the components are -Pi -Pj The force required to neutralise Pi is -Pi and the force required to neutralise Pj is -Pj Thanks from Indranil
July 16th, 2019, 09:20 AM   #9
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Quote:
 Originally Posted by Indranil No The components should be Fsin theta and Fcos theta.
Quote:
 Originally Posted by weirddave No, the components are -Pi -Pj
Kids, kids. You're both right.

$\displaystyle |\vec{F}| = \sqrt{F_x^2 + F_y^2} = \sqrt{(-P)^2 + (-P)^2} = P\sqrt{2}$
$\displaystyle F_x = (P\sqrt{2})\cos(225°) = -P$
$\displaystyle F_y = (P\sqrt{2})\sin(225°) = -P$

Last edited by skipjack; July 17th, 2019 at 12:08 PM.

July 17th, 2019, 12:08 AM   #10
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Quote:
 Originally Posted by DarnItJimImAnEngineer Kids, kids. You're both right
You cheated!

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