# 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.