Vector that describes the power

[mathmari]

Hello!

A mass of 1 kilogram ($1$ kg) that lies at $(0,0,0)$ is hanging from ropes tied at the points $(1,1,1)$ and $(-1,-1,1)$. If the gravitational force has the direction of the vector $\overrightarrow{-k}$ which is the vector that describes the power (voltage) along each of the ropes?

Could you give me some hints?? (Wondering)

 

[I like Serena]

Hello!

A mass of 1 kilogram ($1$ kg) that lies at $(0,0,0)$ is hanging from ropes tied at the points $(1,1,1)$ and $(-1,-1,1)$. If the gravitational force has the direction of the vector $\overrightarrow{-k}$ which is the vector that describes the power (voltage) along each of the ropes?

Could you give me some hints?? (Wondering)

Hey! :)

Power or voltage seem to be unrelated.
Perhaps force was intended? (Wondering)

In that case we have the equilibrium conditions that the sum of the force components along each axis must be zero. (Wasntme)

 

[mathmari]

Perhaps force was intended? (Wondering)

Yes... (Nod)

In that case we have the equilibrium conditions that the sum of the force components along each axis must be zero. (Wasntme)

So, we have too use the following: $$\sum F_i=0 \Rightarrow F_1+F_2+F_3=0$$ right??

But how?? (Wondering)

 

[I like Serena]

So, we have too use the following: $$\sum F_i=0 \Rightarrow F_1+F_2+F_3=0$$ right??

But how?? (Wondering)

That should be:
$$\sum F_x = 0$$
$$\sum F_y = 0$$
$$\sum F_z = 0$$
(Nerd)

Suppose you have some unknown tensional force $F_1$ respectively $F_2$ in the ropes, combined with the known gravitational force $mg$ that points downwards.
How would these equations looks? (Thinking)

 

[mathmari]

That should be:
$$\sum F_x = 0$$
$$\sum F_y = 0$$
$$\sum F_z = 0$$
(Nerd)

Suppose you have some unknown tensional force $F_1$ respectively $F_2$ in the ropes, combined with the known gravitational force $mg$ that points downwards.
How would these equations looks? (Thinking)

Is it as followed??

$$F_{1x}+F_{2x}=0 \\ F_{1y}+F_{2y}=0 \\ F_{1z}+F_{2z}-mg=0$$

(Wondering)

 

[I like Serena]

Is it as followed??

$$F_{1x}+F_{2x}=0 \\ F_{1y}+F_{2y}=0$$

(Wondering)

Yep. (Nod)

$$F_{1z}+F_{2z}-mg=0$$

Not quite.
E.g. $F_{1z}$ should be divided by the length of its direction vector. (Worried)

 

[mathmari]

E.g. $F_{1z}$ should be divided by the length of its direction vector. (Worried)

Why?? (Wondering)

 

[I like Serena]

Why?? (Wondering)

If the tensional force in a rope is $F$ and the directional vector is $(1,1,1)$, then the force vector is $(\frac F{\sqrt 3}, \frac F{\sqrt 3}, \frac F{\sqrt 3})$. (Wasntme)

 

[mathmari]

If the tensional force in a rope is $F$ and the directional vector is $(1,1,1)$, then the force vector is $(\frac F{\sqrt 3}, \frac F{\sqrt 3}, \frac F{\sqrt 3})$. (Wasntme)

Why is the directional vector $(1, 1, 1)$ ?? (Wondering)

The gravitational force has the direction of the vector $-\overrightarrow{k}$.

Does this mean that the directional vector is $(1,1, -1)$ ?? (Wondering)

 

[I like Serena]

Why is the directional vector $(1, 1, 1)$ ?? (Wondering)

Because one of the ropes is attached at (1,1,1) while it is connected to the mass at (0,0,0).
That means that the directional vector is (1,1,1) - (0,0,0) = (1,1,1). (Nerd)

The gravitational force has the direction of the vector $-\overrightarrow{k}$.

Does this mean that the directional vector is $(1,1, -1)$ ?? (Wondering)

The down vector $-\overrightarrow{k}$ is equal to $(0,0,-1)$. (Wasntme)

 

[mathmari]

I tried to make a sketch of the problem...

View attachment 4040

Is this correct?? (Wondering)

Have I drawn the forces $F_1, F_2$, that we are asked to find, correct??

 

[I like Serena]

I tried to make a sketch of the problem...

Is this correct?? (Wondering)

Have I drawn the forces $F_1, F_2$, that we are asked to find, correct??

Yep. Looks good! ;)

 

[mathmari]

Yep. Looks good! ;)

Can we say that since the vector of the force $F_1$ is on the vector $(1, 1, 1)$, it is of the form $F_1(\lambda)=\lambda (1, 1, 1)$ and since the vector of the force $F_2$ is on the vector $(-1, -1, 1)$, it is of the form $F_2(\mu)=\mu (1, 1, 1)$ ?? (Wondering)

 

[I like Serena]

Can we say that since the vector of the force $F_1$ is on the vector $(1, 1, 1)$, it is of the form $F_1(\lambda)=\lambda (1, 1, 1)$ and since the vector of the force $F_2$ is on the vector $(-1, -1, 1)$, it is of the form $F_2(\mu)=\mu (1, 1, 1)$ ?? (Wondering)

Yes to the first. (Nod)
No go the second. (Shake) (Nerd)

 

[mathmari]

Oh... It should be:

Since the vector of the force $F_1$ is on the vector $(1, 1, 1)$, it is of the form $F_1(\lambda)=\lambda (1, 1, 1)$ and since the vector of the force $F_2$ is on the vector $(-1, -1, 1)$, it is of the form $F_2(\mu)=\mu (-1, -1, 1)$

right?? (Wondering)

 

[I like Serena]

Oh... It should be:

Since the vector of the force $F_1$ is on the vector $(1, 1, 1)$, it is of the form $F_1(\lambda)=\lambda (1, 1, 1)$ and since the vector of the force $F_2$ is on the vector $(-1, -1, 1)$, it is of the form $F_2(\mu)=\mu (-1, -1, 1)$

right?? (Wondering)

Right! (Happy)

 

[mathmari]

Is the way I formulated it correct?? Or could I improve something?? (Wondering)

Then we have that $$F_1+F_2+W=0$$ by the equilibrium condition, where $W=9.8 (0, 0, -1)$, right?? (Wondering)

$$\lambda (1, 1, 1)+\mu(-1, -1, 1)+(0, 0, -9.8)=(0, 0, 0) \\ \Rightarrow \lambda-\mu=0 , \ \lambda-\mu=0 , \ \lambda+\mu=9.8 \\ \Rightarrow \lambda=\mu=4.9$$

Is this correct?? (Wondering)

 

[I like Serena]

Is the way I formulated it correct?? Or could I improve something?? (Wondering)

I'd write $F_1=λ(1,1,1)$ instead of $F_1(λ)=λ(1,1,1)$.
That's because $F_1$ is not a function of $λ$. It's a fixed value that has a relationship to the $λ$ that you've just introduced. (Nerd)

Then we have that $$F_1+F_2+W=0$$ by the equilibrium condition, where $W=9.8 (0, 0, -1)$, right?? (Wondering)

$$\lambda (1, 1, 1)+\mu(-1, -1, 1)+(0, 0, -9.8)=(0, 0, 0) \\ \Rightarrow \lambda-\mu=0 , \ \lambda-\mu=0 , \ \lambda+\mu=9.8 \\ \Rightarrow \lambda=\mu=4.9$$

Is this correct?? (Wondering)

Yup! (Nod)

 

[mathmari]

Ok... Thank you! (flower)