MHB Vector that describes the power

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A 1 kg mass hangs at the origin, supported by ropes attached at (1,1,1) and (-1,-1,1), with gravitational force directed downwards. The discussion centers around determining the tension forces in the ropes, using equilibrium conditions for force components along the x, y, and z axes. Participants clarify that the directional vectors for the forces must account for the lengths of the ropes, leading to specific formulations for the tension forces. The final equations derived indicate that both tension forces are equal at 4.9 N, confirming the equilibrium of the system. The conversation emphasizes the importance of correctly representing the forces and their relationships in vector form.
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Hello! :o

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)
 
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mathmari said:
Hello! :o

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)
 
I like Serena said:
Perhaps force was intended? (Wondering)

Yes... (Nod)
I like Serena said:
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)
 
mathmari said:
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)
 
I like Serena said:
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)
 
mathmari said:
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)
 
I like Serena said:
E.g. $F_{1z}$ should be divided by the length of its direction vector. (Worried)

Why?? (Wondering)
 
mathmari said:
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)
 
I like Serena said:
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)
 
  • #10
mathmari said:
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)
 
  • #11
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??
 

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  • #12
mathmari said:
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! ;)
 
  • #13
I like Serena said:
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)
 
  • #14
mathmari said:
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)
 
  • #15
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)
 
  • #16
mathmari said:
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)
 
  • #17
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)
 
  • #18
mathmari said:
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)
 
  • #19
Ok... Thank you! (flower)
 
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