How Do Sign Conventions Affect Equations in Newton's Second Law?

[undividable]

Imagine we have a box suspended to the ceiling by a rope of negligible mass, the net force looks like:
∑F=T-Fg=0
T=Fg
where T and Fg are equal in magnitude but opposite in direction.
If we cut the rope, the box is in free fall and ∑F=Fg, and this equation is similar to T=Fg
but Fg and ∑F have the same direction and T and Fg have opposite directions.
I am really confused about this, and it must come from the sign convention while using vector notation and when you drop the vector notation,
Thanks for the attention :)

 

[Chestermiller]

Your force balance for when the rope is cut is incorrect. Do you really think that there is tension in the rope after you cut it?

 

[undividable]

Your force balance for when the rope is cut is incorrect. Do you really think that there is tension in the rope after you cut it?

like i said " If we cut the rope, the box is in free fall and ∑F=Fg"

 

[Chestermiller]

like i said " If we cut the rope, the box is in free fall and ∑F=Fg"

So, $$ma=F_g=mg$$ where a is the downward acceleration after the rope is cut. So, $a=g$. Does this clear things up?

 

[I like Serena]

Imagine we have a box suspended to the ceiling by a rope of negligible mass, the net force looks like:
∑F=T-Fg=0
T=Fg
where T and Fg are equal in magnitude but opposite in direction.
If we cut the rope, the box is in free fall and ∑F=Fg, and this equation looks like T=Fg
but Fg and ∑F have the same direction.
I am really confused about this, and it must come from the sign convention while using vector notation and when you drop the vector notation,
Thanks for the attention :)

The convention here is that we take positive to be up.
Since the force of tension on the box is up, we pick +T, where T is the magnitude of the force of tension.
And the force of gravity is down, so we pick -Fg, where Fg is the magnitude (that is, a positive number).
So the sum of the forces is ∑F = +T - Fg.

When we cut the rope, we're left with only the force of gravity, and we have, if we're consistent, ∑F = -Fg.
Saying ∑F=Fg is a bit sloppy, since we're neglecting the choice that we've made for positive.
Instead we're picking down to be positive - but just in this particular case.

 

[undividable]

The convention here is that we take positive to be up.
Since the force of tension on the box is up, we pick +T, where T is the magnitude of the force of tension.
And the force of gravity is down, so we pick -Fg, where Fg is the magnitude (that is, a positive number).
So the sum of the forces is ∑F = +T - Fg.

When we cut the rope, we're left with only the force of gravity, and we have, if we're consistent, ∑F = -Fg.
Saying ∑F=Fg is a bit sloppy, since we're neglecting the choice that we've made for positive.
Instead we're picking down to be positive - but just in this particular case.

But if ∑F = -Fg, that would mean that the net force and the gravitational force have different directions, and that is not the case

 

[Chestermiller]

Why don't you do it using unit vectors and see how that works for you?

 

[undividable]

Why don't you do it using unit vectors and see how that works for you?

does it change anything? won't the signs still be the same?

 

[Chestermiller]

Well, let's see what we get. Before the rope is cut, the forces acting on the box are $T\mathbf{i_y}$ and $mg(\mathbf{-i_y})$ So, at equilibrium, the sum of the forces on the box are equal to zero: $$T\mathbf{i_y}+mg(\mathbf{-i_y})=0$$or equivalently:$$T\mathbf{i_y}=mg(\mathbf{i_y})$$So, $$T=mg$$
After the rope is cut, the only force acting on the box is $mg(\mathbf{-i_y})$. This is the sum of the forces acting on the box. So, this must be equal to its mass times its acceleration: $$m\mathbf{a}=mg(\mathbf{-i_y})$$Therefore, $$\mathbf{a}=g(\mathbf{-i_y})=-g\mathbf{i_y}$$This tells us that the acceleration is in the negative y-direction.

Do you now see how using unit vectors helps to settle sign issues.

 

[undividable]

Well, let's see what we get. Before the rope is cut, the forces acting on the box are $T\mathbf{i_y}$ and $mg(\mathbf{-i_y})$ So, at equilibrium, the sum of the forces on the box are equal to zero: $$T\mathbf{i_y}+mg(\mathbf{-i_y})=0$$or equivalently:$$T\mathbf{i_y}=mg(\mathbf{i_y})$$So, $$T=mg$$
After the rope is cut, the only force acting on the box is $mg(\mathbf{-i_y})$. This is the sum of the forces acting on the box. So, this must be equal to its mass times its acceleration: $$m\mathbf{a}=mg(\mathbf{-i_y})$$Therefore, $$\mathbf{a}=g(\mathbf{-i_y})=-g\mathbf{i_y}$$This tells us that the acceleration is in the negative y-direction.

Do you now see how using unit vectors helps to settle sign issues.

but why didn't you use unit vector on ma? in ma=mg(-iy), isn't the net force also a vector?

 

[Chestermiller]

but why didn't you use unit vector on ma? in ma=mg(-iy), isn't the net force also a vector?

At that point, we didn't know what the direction of a should be. We used this analysis to determine that.

 

[Dale]

Imagine we have a box suspended to the ceiling by a rope of negligible mass, the net force looks like:
∑F=T-Fg=0
T=Fg

So I think that your confusion about signs later starts with some sloppy notation here. The net force is a vector, so T and Fg should be vectors also. In which case it is
∑F=T+Fg=0
T=-Fg
Where all of the quantities above are vectors. Since by definition we know that the Fg vector points downwards, then this shows us that T points upwards.

Now, we can choose to adopt coordinates where the positive z direction is vertical upwards. Then, we could write the z component of Newton's 2nd law as
∑Fz=Tz-Fgz=0
Where all these quantities are components of the vectors so T=(0,0,Tz) and Fg=(0,0,-Fgz)

But if ∑F = -Fg, that would mean that the net force and the gravitational force have different directions, and that is not the case

So, now having written our notation more clearly we would either have the vector equation
∑F=Fg
Or the component equation
∑Fz=-Fgz
The first indicates that the net force is in the same direction as gravity (downward) and the second indicates that the z component of the net force is negative (downward). Both are correct and both say the same thing, but one is a vector equation and the other is just the z component.

It is important to keep your notation tidy and to properly distinguish between vectors and components. Also, when you are using coordinates you need to keep in mind what coordinate system you have chosen and be consistent.

 

[undividable]

So I think that your confusion about signs later starts with some sloppy notation here. The net force is a vector, so T and Fg should be vectors also. In which case it is
∑F=T+Fg=0
T=-Fg
Where all of the quantities above are vectors. Since by definition we know that the Fg vector points downwards, then this shows us that T points upwards.

Now, we can choose to adopt coordinates where the positive z direction is vertical upwards. Then, we could write the z component of Newton's 2nd law as
∑Fz=Tz-Fgz=0
Where all these quantities are components of the vectors so T=(0,0,Tz) and Fg=(0,0,-Fgz)

So, now having written our notation more clearly we would either have the vector equation
∑F=Fg
Or the component equation
∑Fz=-Fgz
The first indicates that the net force is in the same direction as gravity (downward) and the second indicates that the z component of the net force is negative (downward). Both are correct and both say the same thing, but one is a vector equation and the other is just the z component.

It is important to keep your notation tidy and to properly distinguish between vectors and components. Also, when you are using coordinates you need to keep in mind what coordinate system you have chosen and be consistent.

∑F=T+Fg=0
T=-Fg
Where all of the quantities above are vectors

substituting the vector components here we get Tz=Fgz, right? this component equation is the same as yours ∑Fz=Tz-Fgz=0

∑Fz=Tz-Fgz=0
Where all these quantities are components of the vectors so T=(0,0,Tz) and Fg=(0,0,-Fgz)

but in the following step you did things differently

So, now having written our notation more clearly we would either have the vector equation
∑F=Fg
Or the component equation
∑Fz=-Fgz

so in the vector equation ∑F=Fg, , as in the vector equation T=-Fg, when we put in the vector components we get -∑F=-Fg, as in when we put the vecotr component in T=-Fg we get T=Fg, but you said we get ∑Fz=-Fgz in the component equation of ∑F=Fg, i am really confused :( because i know that -∑F=-Fg is wrong because that would mean the acceleration points in the positive direction , (upwards) which i know is wrong

 

[Chestermiller]

If you do it using unit vectors, as I did in post #9, you can't go wrong.

 

[undividable]

If you do it using unit vectors, as I did in post #9, you can't go wrong.

ma=mg(−iy)

shouldn't it be ma(-iy)=mg(-iy) since the net force and the gravitational force are in the negative direction?, as in T(iy)=Fg(-iy), where they are in opposite directions

 

[Chestermiller]

shouldn't it be ma(-iy)=mg(-iy) since the net force and the gravitational force are in the negative direction?, as in T(iy)=Fg(-iy), where they are in opposite directions

$\mathbf{a}$ is a vector, as indicated in boldface. But we don't know its direction yet. So we are leaving its direction as yet unspecified. We are using the force balance equation $m\mathbf{a}=mg(\mathbf{-i_y})$ to determine the direction of the acceleration vector $\mathbf{a}$. This force balance equation tells us that $\mathbf{a}$ is in the same direction as $-\mathbf{i_y}$ (i.e., downward).

Incidentally, your equation involving T is incorrect. It should read:
$$\mathbf{T}+F_g(\mathbf{-i_y})=\mathbf{0}$$or equivalently$$\mathbf{T}=F_g\mathbf{i_y}$$

 

[Dale]

substituting the vector components here we get Tz=Fgz, right?

Yes.

this component equation is the same as yours ∑Fz=Tz-Fgz=0

Yes.

so in the vector equation ∑F=Fg,

After cutting the rope, yes.

, as in the vector equation T=-Fg,

Before we cut the rope, yes

when we put in the vector components we get -∑F=-Fg,

No, this is the vector equation (after cutting the rope), not the component equation.

Why did you put the negative sign on both sides? Most people would cancel them out.

we get T=Fg,

No, T=-Fg (vector equation before cutting the rope)

you said we get ∑Fz=-Fgz

Yes (after cutting the rope)

in the component equation of ∑F=Fg,

Again, this is a vector equation, not a component equation. I defined Fg=(0,0,-Fgz), with the z unit vector pointing upwards. So Fgz is a positive number. I think you may have missed that.

i know that -∑F=-Fg is wrong

No, it is correct (after cutting the rope). The only force is gravity so the net force is equal to the gravitational force by definition.

that would mean the acceleration points in the positive direction , (upwards)

I am not sure why you think that.