How to Calculate Torque of a Leaning Rigid Body

1. Jun 12, 2012

Bennigan88

So this is a conceptual question, it's not a direct homework question, but it does involve how to do a kind of calculation. I hope this isn't the wrong place to post a nonspecific question like this. In the case of a board/stick/ladder leaning against a wall, about an axis O at the bottom of the object where it is in contact with the ground, why is the distance used to determine torque (which is Force x Distance x sin θ) halfway up the ladder? Ladder is 15m long in my example.

I tried to use integration to calculate the torque, and this is what I ended up with:

$$\tau = mg \times x \times \sin\theta = mgsin\theta \cdot x \\ d\tau = mg \sin\theta \cdot dx \\ \int d\tau = \int mgsin\theta dx \\ \int d\tau = mg \sin\theta \int dx \\ = mg \sin\theta \int_0^{15} dx \\ = mgsin\theta \cdot x |_0^{15} \\ = 15 \cdot mg \sin\theta$$

So why is the scalar for distance 1/2 of the length rather than the entire length like my calculations? What am I missing? Thanks for any insight.

!! The mass also changes, so I think I need a differential mass element... more thinking required :/

Last edited: Jun 13, 2012
2. Jun 13, 2012

Simon Bridge

Because gravity acts at the center of mass.

3. Jun 13, 2012

Bennigan88

Or can I use a Riemann sum to calculate the torque? Assume the ladder is divided up into n pieces, and the mass of each piece is its linear density λ times the change in distance.

$$\tau = x \cdot F \cdot \sin\theta \\ F = m \cdot g \\ m = \lambda \Delta x \\ \tau_i = x_i \cdot \lambda \Delta x \cdot g \cdot \sin\theta = \lambda g \sin\theta \cdot x_i \cdot \Delta x \\ \sum_i^n \lambda g \sin\theta \cdot x_i \cdot \Delta x \\ \lim_{n\to\infty}\sum_i^n \lambda g \sin\theta \cdot x_i \cdot \Delta x = \int_0^{15} \lambda g \sin\theta x dx\\ = \lambda g \sin\theta \cdot \int_0^{15} x dx = \lambda g \sin\theta \cdot \left[ \dfrac{1}{2}x^2 \right]_0^{15}$$

I don't see this moving towards a validation of the torque being calculated at the half-way point with mg acting on the center of gravity... can anyone nudge me in the right direction?

4. Jun 13, 2012

Bennigan88

I am aware of that, but shouldn't I end up finding the same torque using integration?

5. Jun 13, 2012

Simon Bridge

Well yes - and the integrated part will amount to finding the center of mass.
So your question amounts to: "how come I keep blowing the math?"

6. Jun 13, 2012

haruspex

Not sure what the first equation is saying, but the second is wrong. The RHS is the mass of the element dx multiplied by g sin(θ). The distance is x sin(θ):
$$d\tau = mgxsin\theta \cdot dx \\$$

7. Jun 13, 2012

Simon Bridge

* Actually I think you got it right in the reiman sum - the first integral was poorly set up which is why it didn't work.

Finish the calculation.
$$g\lambda\sin(\theta)\int_0^L x.dx = g\lambda\sin(\theta)\cdot \frac{1}{2}x^2 \bigg |_0^L = \frac{g\lambda L^2}{2} \sin(\theta)$$
... which is what you want - because λL = m, the mass of the ladder.

In the first derivation, you misidentified the elements - it should have gone from $\tau = mgx\sin(\theta)$ to $d\tau = gx\sin(\theta)\cdot dm$.
In the second derivation you lost confidence just before the payoff :)

Last edited: Jun 13, 2012
8. Jun 13, 2012

Bennigan88

You're right! I'm thrilled!!! I need to get better at recognizing things in the answers. This isn't the first time I've gotten something more or less right but didn't see it because I didn't make the connection. Thanks for seeing it through with me! Generalizing the 15 to L was I think the coup de grace that I was missing.