Horizontal Tension Force equation

[karen_lorr]

TL;DR

Horizontal Tension Force equation between two equal (pulling) forces

(not a homework question - I left education over 45 years ago)

Hi, this just out of interest as I have been googleing this and can't find the answer

I know the equation for tension force, for example a weight on a string, block on a ramp, etc.

But is there something similar for a horizontal pull (on a cable, string, etc) between two sides of an equal system. For example, a belt buckle – I know silly example or even a string between your outstretched arms.

Let’s say you (somehow) attached a piece of cotton to the petals of a rose. In the morning the rose would open. The petals would move apart. The cotton would have a tension (pull between the petals). Even if one petal was larger (more mass, more pulling power, etc) the tension on the cotton would be the same at both ends.

Is there a standard formula for this?

I hope this make sense.

Thank you

 

[PeroK]

If you hold a piece of string horizontally and pull at both ends, then the tension depends on how hard you pull.

I can't imagine rose petals could generate much pulling force, but then horticulture isn't my strong suit!

 

[karen_lorr]

You’re probably right about the rose.

But let’s say each petal opens from the centre with a force of 123kn (very strong roses).

There are two petals each pulling with the same force.

The cotton is not massless (as infinite acceleration hurts my brain), so let’s say it has a weight which is so small it hardy affects the outcome.

The gravity on each petal is the same as the normal force, so is cancelled.

Is there a formula for this type of action?

 

[PeroK]

You’re probably right about the rose.

But let’s say each petal opens from the centre with a force of 123kn (very strong roses).

There are two petals each pulling with the same force.

The cotton is not massless (as infinite acceleration hurts my brain), so let’s say it has a weight which is so small it hardy affects the outcome.

The gravity on each petal is the same as the normal force, so is cancelled.

Is there a formula for this type of action?

The tension in a rope or string is equal to the force applied at either end. That's where there is equilibrium.

 

[jack action]

I know the equation for tension force, for example a weight on a string, block on a ramp, etc.

I guess the equation you know is $F_g=mg$.

Tension is basically a force ($F$). The previous equation is a special case for a mass ($m$) where the force due to gravity ($F_g$) would cause an equivalent acceleration $g$, if unopposed by another force.

But this equation is a special case of a more general equation:
$$\sum F = ma$$
Where $\sum F$ is the summation of all forces in one direction and $a$ is the acceleration of a mass $m$ in the same direction.

But it is possible to not have an acceleration, i.e. $a=0$, therefore $ma = 0$ and thus $\sum F = 0$.

That last equation means that there are at least 2 forces that are equal and opposite. For example:
$$F_1 + \left(-F_2\right) = 0$$
$F_2$ has a negative sign to show that it goes in the opposite direction of $F_1$. So this equation basically states that $F1$ and $F_2$ must have the same magnitude, the same numeric value. If each one of these forces are acting on opposite end of a string, the string will be under a tension of magnitude $F_1$ (or $F_2$, since they have the same value).

If $F_1$ and $F_2$ are of different magnitudes, then the acceleration $a$ cannot be zero. Therefore the string would begin to move. The sign of the acceleration would dictate the direction of the acceleration. Note that if accelerations are not involved, the mass is irrelevant.

Let’s say you (somehow) attached a piece of cotton to the petals of a rose. In the morning the rose would open. The petals would move apart. The cotton would have a tension (pull between the petals). Even if one petal was larger (more mass, more pulling power, etc) the tension on the cotton would be the same at both ends.

Is there a standard formula for this?

Since we're looking at the horizontal forces only, the previous equation applies where each petal pulls one against each other.

But let’s say each petal opens from the centre with a force of 123kn (very strong roses).

The tension in the piece of cotton would be 123 kN ($=F_1 = F_2$).

The cotton is not massless (as infinite acceleration hurts my brain), so let’s say it has a weight which is so small it hardy affects the outcome.

Mass is irrelevant in this case as the acceleration would be zero.

The gravity on each petal is the same as the normal force, so is cancelled.

The gravity is irrelevant here since it acts only in the vertical plane, which is perpendicular to the plane we are looking at in the present case.

But we can analyze this vertical plane also where $F_g = ma$ or $mg = ma$. This would mean that the piece of cotton is under acceleration ($a=g$) and should slide down the petals. If it is not the case, then it is because there is another force involved, in this case, most likely friction $F_f$. The equation is then:
$$F_g + \left(-F_f\right) = m(0)$$
Or:
$$ F_g = F_f$$
So the friction force is equal and opposite to the gravity and the acceleration is zero.

 

[Stephen Tashi]

Even if one petal was larger (more mass, more pulling power, etc) the tension on the cotton would be the same at both ends.

If the string is acted upon only by two unequal forces at its ends, it wouldn't remain stationary. By analogy, in a tug-of-war, the stronger side wins by causing the rope to move.

Were you expecting to find a formula for the tension on a string acted upon only by unequal forces at each end? A net force in one direction would act on each section of such a string. That net force wouldn't be called a "tension" since the word "tension" implies a balance of forces.