MHB How can we find the formula of the force?

[evinda]

Hello! (Wave)

A force $6N$ forms an angle $\frac{\pi}{4}$ with the y-axis, showing to the right.
The force is opposed by the movement of an object along the line that connects $(1,2)$ with $(5,4)$.
I want to find a formula for the vector-force $F$.

The line that connects $(1,2)$ with $(5,4)$ is $l(t)=(1,2)+t(4,2), t \in \mathbb{R}$.
How can we find the formula of the force?

 

[I like Serena]

Hello! (Wave)

A force $6N$ forms an angle $\frac{\pi}{4}$ with the y-axis, showing to the right.
The force is opposed by the movement of an object along the line that connects $(1,2)$ with $(5,4)$.
I want to find a formula for the vector-force $F$.

The line that connects $(1,2)$ with $(5,4)$ is $l(t)=(1,2)+t(4,2), t \in \mathbb{R}$.
How can we find the formula of the force?

Hey evinda! (Smile)

Forces and movements live in different universes.
It's because of a force that a movement changes, but movements have no impact on forces whatsoever.
Apparently your force is $\mathbf F = (6/\sqrt 2, 6/\sqrt 2) \text{ N}$. The movement does not change that. (Nerd)

 

[evinda]

Hey evinda! (Smile)

Forces and movements live in different universes.
It's because of a force that a movement changes, but movements have no impact on forces whatsoever.
Apparently your force is $\mathbf F = (6/\sqrt 2, 6/\sqrt 2) \text{ N}$. The movement does not change that. (Nerd)

Why does it hold that $\mathbf F = (6/\sqrt 2, 6/\sqrt 2) \text{ N}$ ? How do we deduce it? (Thinking)

 

[I like Serena]

Why does it hold that $\mathbf F = (6/\sqrt 2, 6/\sqrt 2) \text{ N}$ ? How do we deduce it? (Thinking)

An angle of $\frac\pi 4$ with the y-axis showing to the right corresponds with the unit vector $\frac 1 {\sqrt 2} (1,1)$.
Multiply by the value of the force to get the force vector. (Nerd)

 

[evinda]

An angle of $\frac\pi 4$ with the y-axis showing to the right corresponds with the unit vector $\frac 1 {\sqrt 2} (1,1)$.

Why is it like that? Could you explain it further to me? (Thinking)

 

[I like Serena]

Why is it like that? Could you explain it further to me? (Thinking)

The unit vector along the y-axis is $(0,1)$.
Suppose $(u,v)$ is a unit vector that makes an angle of $\frac \pi 4$ with it.
Then their dot product has to be:
$$(u,v) \cdot (0,1) = 1 \cdot 1 \cdot \cos \frac \pi 4$$
So:
$$v = \frac 1 {\sqrt 2}$$
And since $\sqrt{u^2+v^2} = 1$:
$$u = \pm \frac 1 {\sqrt 2}$$
(Thinking)

 

[evinda]

The unit vector along the y-axis is $(0,1)$.
Suppose $(u,v)$ is a unit vector that makes an angle of $\frac \pi 4$ with it.
Then their dot product has to be:
$$(u,v) \cdot (0,1) = 1 \cdot 1 \cdot \cos \frac \pi 4$$
So:
$$v = \frac 1 {\sqrt 2}$$
And since $\sqrt{u^2+v^2} = 1$:
$$u = \pm \frac 1 {\sqrt 2}$$
(Thinking)

Do we have to suppose that $(u,v)$ shows to the right to reject $u=-\frac{1}{\sqrt{2}}$?

 

[I like Serena]

Do we have to suppose that $(u,v)$ shows to the right to reject $u=-\frac{1}{\sqrt{2}}$?

Yep. (Nod)

 

[evinda]

The unit vector along the y-axis is $(0,1)$.

Could we also pick the unit vector along the y-axis $(0,-1)$? Or does $v$ have to be positive?

 

[I like Serena]

Could we also pick the unit vector along the y-axis $(0,-1)$? Or does $v$ have to be positive?

That's a unit vector along the negative y-axis.
A unit vector along the y-axis is considered to be along the positive y-axis. (Nerd)

 

[evinda]

Yep. (Nod)

View attachment 5326

If we would take $u=-\frac{1}{\sqrt{2}}$ the vector would show to the left as at the above graph, right?

 

[I like Serena]

If we would take $u=-\frac{1}{\sqrt{2}}$ the vector would show to the left as at the above graph, right?

Right. (Nod)

 

[evinda]

Right. (Nod)

Great!
I want to find also the angle $\theta$ that form the vector of the direction of the movement $D=(5-1)i+(4-2)j$ and the direction of the force $F$.

I have thought the following:

$$F \cdot D=||F|| ||D|| \cos{\theta} \Rightarrow \left( \frac{6}{\sqrt{2}},\frac{6}{\sqrt{2}} \right) \cdot (4,2)=6 \sqrt{20} \cos{\theta} \Rightarrow \cos{\theta}=\frac{3}{\sqrt{10}}$$Am I right?

 

[I like Serena]

Great!
I want to find also the angle $\theta$ that form the vector of the direction of the movement $D=(5-1)i+(4-2)j$ and the direction of the force $F$.

I have thought the following:

$$F \cdot D=||F|| ||D|| \cos{\theta} \Rightarrow \left( \frac{6}{\sqrt{2}},\frac{6}{\sqrt{2}} \right) \cdot (4,2)=6 \sqrt{20} \cos{\theta} \Rightarrow \cos{\theta}=\frac{3}{\sqrt{10}}$$Am I right?

Yep! (Nod)

 

[evinda]

Yep! (Nod)

Nice... How can I find the value of $\theta$ in radians? (Thinking)

 

[I like Serena]

Nice... How can I find the value of $\theta$ in radians? (Thinking)

That's
$$\theta = \arccos {\frac 3 {\sqrt {10}}}$$
There doesn't seem to be a nicer expression for that. (Worried)

 

[evinda]

That's
$$\theta = \arccos {\frac 3 {\sqrt {10}}}$$
There doesn't seem to be a nicer expression for that. (Worried)

Nice... Thanks a lot! (Cool)

 

[evinda]

If we would have the following:The work $W$ that is produced if an object moves from $(0,0)$ to $(7,2)$ by the influence of a force $F$ is $W=F \cdot r$, where $r$ is a vector with starting point $(0,0)$ and end point $(7,2)$. We suppose that the force $F$ has length $6 lb$ and forms an angle $\frac{\pi}{6} rad$ with the horizontal axis with direction to the right.

I want to find $W$.

Again I took a unit vector $(u,v)$ that forms angle $\frac{\pi}{6}$ with the $x$-axis and shows to the right.
A unit vector along the $x$-axis is $(1,0)$.

Then using the dot-product, we get that $u=\frac{\sqrt{3}}{2}$.

Then since $(u,v)$ is a unit vector we get $v=\pm \frac{1}{2}$.

How do we reject one of $\frac{1}{2}$ or $-\frac{1}{2}$ ?

Wouldn't the vector show at both cases to the right?

 

[I like Serena]

If we would have the following:The work $W$ that is produced if an object moves from $(0,0)$ to $(7,2)$ by the influence of a force $F$ is $W=F \cdot r$, where $r$ is a vector with starting point $(0,0)$ and end point $(7,2)$. We suppose that the force $F$ has length $6 lb$ and forms an angle $\frac{\pi}{6} rad$ with the horizontal axis with direction to the right.

I want to find $W$.

Again I took a unit vector $(u,v)$ that forms angle $\frac{\pi}{6}$ with the $x$-axis and shows to the right.
A unit vector along the $x$-axis is $(1,0)$.

Then using the dot-product, we get that $u=\frac{\sqrt{3}}{2}$.

Then since $(u,v)$ is a unit vector we get $v=\pm \frac{1}{2}$.

How do we reject one of $\frac{1}{2}$ or $-\frac{1}{2}$ ?

Wouldn't the vector show at both cases to the right?

Hey evinda! (Smile)

A positive angle is conventionally counter clock wise, that is, upward.
At least that's what I would assume without any other information.

 

[evinda]

Do we use the fact that the force $F$ forms an angle $\frac{\pi}{6} rad$ with the horizontal axis with direction to the right ? If so, at which point?

 

[I like Serena]

Do we use the fact that the force $F$ forms an angle $\frac{\pi}{6} rad$ with the horizontal axis with direction to the right ? If so, at which point?

A vector of itself has no position - there is no point.
We can pick any point we'd like, say, the origin.
Then we might say that it makes an angle of $\frac{\pi}{6} \text{ rad}$ on the unit circle.
Or that it makes an angle of $\frac{\pi}{6} \text{ rad}$ with the positive x-axis.
More generally, it's with any horizontal line with direction to the right. (Nerd)