Vector Algebra: Finding Resultant Forces at Optimal Angles | Expert Help

[IonizingJai]

Question: At what angles must be the two forces $\vec A+\vec B$ and $\vec A-\vec B$ act so that the resultant may be :
$$\sqrt{ A^2+B^2}$$
Attempt at solution :
Let the given forces be $\vec F_1=\vec A+\vec B$ and $\vec F_2=\vec A-\vec B$ .
Now, Resultant vector : $\vec F_1 + \vec F_2 = (\vec A+\vec B) + (\vec A-\vec B)$ .
Magnitude of Resultant: $|\vec F_1 + \vec F_2| = \sqrt{(A+B)^2+(A-B)^2+2(A+B)(A-B)\cos(\theta)}$.
Here, $\theta$ being the angle between $\vec F_1 and \vec F_2$ .
Stuck here $=\sqrt{2(A^2+B^2)+2(A^2-B^2)\cos(\theta)}$ . [ I'm noob with LaTeX too.].

 

[HallsofIvy]

Question: At what angles must be the two forces $\vec A+\vec B$ and $\vec A-\vec B$ act so that the resultant may be :
$$\sqrt{ A^2+B^2}$$
Attempt at solution :
Let the given forces be $\vec F_1=\vec A+\vec B$ and $\vec F_2=\vec A-\vec B$ .
Now, Resultant vector : $\vec F_1 + \vec F_2 = (\vec A+\vec B) + (\vec A-\vec B)$ .

You realize that $(\vec{A}+ \vec{B})+ (\vec{A}- \vec{B})= 2\vec{A}$ don't you? And, since the magnitude of $2\vec{A}$ does not depend on $\vec{B}$, I have to ask "the resultant of what"?
You seem to have misunderstood the question.

Magnitude of Resultant: $|\vec F_1 + \vec F_2| = \sqrt{(A+B)^2+(A-B)^2+2(A+B)(A-B)\cos(\theta)}$.
Here, $\theta$ being the angle between $\vec F_1 and \vec F_2$ .
Stuck here $=\sqrt{2(A^2+B^2)+2(A^2-B^2)\cos(\theta)}$ . [ I'm noob with LaTeX too.].

 

[Ray Vickson]

Question: At what angles must be the two forces $\vec A+\vec B$ and $\vec A-\vec B$ act so that the resultant may be :
$$\sqrt{ A^2+B^2}$$
Attempt at solution :
Let the given forces be $\vec F_1=\vec A+\vec B$ and $\vec F_2=\vec A-\vec B$ .
Now, Resultant vector : $\vec F_1 + \vec F_2 = (\vec A+\vec B) + (\vec A-\vec B)$ .
Magnitude of Resultant: $|\vec F_1 + \vec F_2| = \sqrt{(A+B)^2+(A-B)^2+2(A+B)(A-B)\cos(\theta)}$.
Here, $\theta$ being the angle between $\vec F_1 and \vec F_2$ .
Stuck here $=\sqrt{2(A^2+B^2)+2(A^2-B^2)\cos(\theta)}$ . [ I'm noob with LaTeX too.].

Do not start a new thread for the same question already asked.

 

[Mark44]

Do not start a new thread for the same question already asked.

It's OK, Ray. I asked him to start a new thread, as I locked and will soon delete the old thread.

 

[IonizingJai]

You realize that (A⃗ +B⃗ )+(A⃗ −B⃗ )=2A⃗ (\vec{A}+ \vec{B})+ (\vec{A}- \vec{B})= 2\vec{A} don't you? And, since the magnitude of 2A⃗ 2\vec{A} does not depend on B⃗ \vec{B}, I have to ask "the resultant of what"?
You seem to have misunderstood the question.

Yeah , i never approached the problem this way, sorry.
Also i found another way :
Let the given forces be $\vec F_1=\vec A+\vec B$ and $\vec F_2=\vec A-\vec B$ .
Now, Resultant vector $\vec F_1 + \vec F_2 = (\vec A+\vec B) + (\vec A-\vec B)$
Magnitude of Resultant:
$|\vec F_1 + \vec F_2| = \sqrt{(A+B)^2+(A-B)^2+2(A+B)(A-B)\cos(\theta)}$ ......1
But according to question the Resultant may be (but it should have been magnitude of the resultant, the question btw is correct as i read it.)
$= \sqrt{(A^2+B^2)}$ .....2
Equating 1 and 2 we get .
and $2(A^2+B^2)+2(A^2-B^2)\cos\theta = (A^2+B^2)$ .
and $2(A^2-B^2)\cos\theta = - (A^2+B^2)$ .
and $\cos\theta= -1/2$.
That is $\theta = cos^{-1}(-0.5)$
$\theta = - 120 degrees$ between the two Forces $\vec F_1 and \vec F_2$ .

 

[Mark44]

Yeah , i never approached the problem this way, sorry.
Also i found another way :
Let the given forces be $\vec F_1=\vec A+\vec B$ and $\vec F_2=\vec A-\vec B$ .
Now, Resultant vector $\vec F_1 + \vec F_2 = (\vec A+\vec B) + (\vec A-\vec B)$

Simplify what's above!

You're making this much harder than it needs to be.

Magnitude of Resultant:
$|\vec F_1 + \vec F_2| = \sqrt{(A+B)^2+(A-B)^2+2(A+B)(A-B)\cos(\theta)}$ ......1
But according to question the Resultant may be (but it should have been magnitude of the resultant, the question btw is correct as i read it.)
$= \sqrt{(A^2+B^2)}$ .....2
Equating 1 and 2 we get .
and $2(A^2+B^2)+2(A^2-B^2)\cos\theta = (A^2+B^2)$ .
and $2(A^2-B^2)\cos\theta = - (A^2+B^2)$ .
and $\cos\theta= -1/2$.
That is $\theta = cos^{-1}(-0.5)$
$\theta = - 120 degrees$ between the two Forces $\vec F_1 and \vec F_2$ .

 

[IonizingJai]

Alright , then that means i am wrong. So , if the question is correct , it ask for the value of the angle between the two forces so that the Resultant maybe
$\sqrt{(A^2+B^2)}$.
Mark44 : as you said i should simplify the but HallsofIvy already showed its not possible to arrive on the answer, since it will equal $2\vec A$ ?
Help ?
BTW , The question as asked in the First post is exactly as it is in language and i just copied it there.

 

[Mark44]

So (A + B) + (A - B) = 2A, right?
What's the magnitude of the sum of the two vectors (which would be the magnitude of the resultant of A + B and A - B)?

 

[IonizingJai]

$A\sqrt{2}$

 

[theodoros.mihos]

Is this correct? $$ \frac{0}{3} = \frac{0}{5} \Rightarrow \,\text{beacuse}\, 0=0 \,\text{then}\, 3=5 $$
If you want:
$$ |\mathbf{F}_1+\mathbf{F}_2| = 2A = \sqrt{A^2+B^2} \Rightarrow 3A^2 = B^2 $$
that mean B²=3A² and A is what you like or vs.

 

[mooncrater]

Will ( A vector) +(B vector) have a magnitude (|A|+|B|)²?

 

[HallsofIvy]

$A\sqrt{2}$

No, that is still wrong! Assuming that you have stated the problem correctly, so that A+ B+ A- B= 2A, the length of 2A is NOT \sqrt{2} times the length of A.

 

[theodoros.mihos]

This is the initial problem: the "resultant" of two vectors A and B must be $\sqrt{A^2+B^2}$.
The writer works on |resutlant|.
The other posibillity is that mean:
$$ \sqrt{A^2+B^2} = (\mathbf{A}+\mathbf{B})\cdot(\mathbf{A}-\mathbf{B}) = A^2 + B^2 +2AB\cos{\theta} \Rightarrow \cos{\theta} = \frac{\sqrt{A^2+B^2}-(A^2+B^2)}{2AB} $$
what I can say?

 

[IonizingJai]

Sorry , but i don't get what are you guys implying ?

 

[vela]

Yeah , i never approached the problem this way, sorry.
Also i found another way :
Let the given forces be $\vec F_1=\vec A+\vec B$ and $\vec F_2=\vec A-\vec B$ .
Now, Resultant vector $\vec F_1 + \vec F_2 = (\vec A+\vec B) + (\vec A-\vec B)$
Magnitude of Resultant:
$|\vec F_1 + \vec F_2| = \sqrt{(A+B)^2+(A-B)^2+2(A+B)(A-B)\cos(\theta)}$

This isn't correct. For example, $\lvert \vec F_1 \rvert^2$ isn't equal to $A^2 + B^2$, and $\lvert \vec F_1 \rvert$ isn't equal to $A+B$. Rather, it should be
$$\lvert \vec F_1 \rvert^2 = \lvert \vec{A} + \vec{B} \rvert^2 = A^2 + B^2 + 2AB\cos\theta_\text{AB},$$ and $\lvert \vec F_1 \rvert$ is the square root of that. You might get the feeling this approach is going to be really messy, and you'd be right. Use Mark's suggestion to simplify first.

 

[IonizingJai]

Thanks for replying vela,
but if you check my solution at post NO , #5 , i have done exactly what you said , i have only left the steps in which i had to cancel many things out and expand and stuff. if you check it you will find that i have done.
and i don't get what Mark44 is implying , the question gets stuck that way since, $\vec A$ will not depend on $\vec B$ and thus there will be no angle between them ?

 

[vela]

Reread what I wrote a bit more carefully. You should see that your expression for the magnitude of the resultant is incorrect.

By the way, I suspect you used $\sqrt{A^2+B^2} = A+B$. That's wrong too.