Why Does Hooke's Law Use a Negative Sign in Scalar Form?

[Heisenberg7]

Assume that we have a block connected to a spring. Also, assume that there is no friction, the spring is massless and ideal. If we were to pull on the block with some force $\vec{F_{pull}}$, we are going to get the spring force $\vec{F_{s}}$ in the opposite direction. Assume that we are moving the block at a constant velocity. Let's now write the Newton's second law for the block: $$\vec{F_{pull}} + \vec{F_{s}} = m\vec{a_x} \implies \vec{F_{pull}} + \vec{F_{s}} = 0 \implies F_{pull} - F_{s} = 0 \implies F_{pull} - (-kx) = 0 \implies x = - \frac{F_{pull}}{k}$$ Now, here comes the problem. $x$ should be positive, not negative (right side of the 0; look at the picture). By, Hooke's law we have that $\vec{F_{s}} = -k\vec{x}$. This makes sense in the vector notation, but after converting it to the scalar notation, why do we need to preserve the minus(it just messes up the numbers)?

 

[berkeman]

Please rewrite this question using vectors and a coordinate system. Are you familiar yet with how to do that?

 

[Heisenberg7]

Please rewrite this question using vectors and a coordinate system. Are you familiar yet with how to do that?

Uhm, I'm sorry, I'm not sure I understand what you are asking me to do. I mean, I've written the equations using vectors. Do you mean like, dividing them into components?

 

[berkeman]

Uhm, I'm sorry, I'm not sure I understand what you are asking me to do. I mean, I've written the equations using vectors. Do you mean like, dividing them into components?

$\overrightarrow{F_{pull}}$, we are going to get the spring force $\overrightarrow{F_{s}}$ in the opposite direction. Assume that we are moving the block at a constant velocity. Let's now write the Newton's second law for the block: $$\overrightarrow{F_{pull}} + \overrightarrow{F_{s}} = m\overrightarrow{a_x} \implies \overrightarrow{F_{pull}} + \overrightarrow{F_{s}} = 0 \implies F_{pull} - F_{s} = 0 \implies F_{pull} - (-kx) = 0 \implies x = - \frac{F_{pull}}{k}$$ Now, here comes the problem. $x$ should be positive, not negative (right side of the 0; look at the picture). By, Hooke's law we have that $\overrightarrow{F_{s}} = -k\overrightarrow{x}$.

Use the bold vector notation in LaTeX instead of the over-arrow, and be consistent with the vector notation through your work. Why did your attempted vector notation disappear as your equation moved along?

 

[Heisenberg7]

Use the bold vector notation in LaTeX instead of the over-arrow, and be consistent with the vector notation through your work. Why did your attempted vector notation disappear as your equation moved along?

Oh, okay. The reason why it disappeared is because I simply summed up the magnitudes of the vectors, or in this case, it's gonna be a minus.

 

[berkeman]

$$\vec F = m \vec a$$
$$\vec F = \vec F_s + \vec F_p$$
$$\vec F = -F_s \hat x + F_p \hat x$$
and so on...

 

[kuruman]

Assume that we have a block connected to a spring. Also, assume that there is no friction, the spring is massless and ideal. If we were to pull on the block with some force $\vec{F_{pull}}$, we are going to get the spring force $\vec{F_{s}}$ in the opposite direction. Assume that we are moving the block at a constant velocity. Let's now write the Newton's second law for the block: $$\vec{F_{pull}} + \vec{F_{s}} = m\vec{a_x} \implies \vec{F_{pull}} + \vec{F_{s}} = 0 \implies F_{pull} - F_{s} = 0 \implies F_{pull} - (-kx) = 0 \implies x = - \frac{F_{pull}}{k}$$ Now, here comes the problem. $x$ should be positive, not negative (right side of the 0; look at the picture). By, Hooke's law we have that $\vec{F_{s}} = -k\vec{x}$. This makes sense in the vector notation, but after converting it to the scalar notation, why do we need to preserve the minus(it just messes up the numbers)?

When you write
$\vec{F_{pull}} + \vec{F_{s}} = 0 \implies F_{pull} - F_{s} = 0$, you are replacing the vectors with their magnitudes and you put a negative sign in them when they are both on the same side of the equation. It is the same as saying $F_{pull} = F_{s}$ or $10 =10$. That's OK so far.

What you cannot do is say that the positive number $F_s$ is the same as the negative number $-kx.$ What you can say is that $F_s=k|x|.$

 

[Chestermiller]

The tension in the spring should be $T=k\Delta L$ where $\Delta L$ is the change in length relative to the unextended configuration. The tension always pulls in the direction away from the block.