How to Calculate Torque of a Leaning Rigid Body

[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:

<br /> \tau = mg \times x \times \sin\theta = mgsin\theta \cdot x \\<br /> <br /> d\tau = mg \sin\theta \cdot dx \\<br /> <br /> \int d\tau = \int mgsin\theta dx \\<br /> <br /> \int d\tau = mg \sin\theta \int dx \\ = mg \sin\theta \int_0^{15} dx \\<br /> <br /> = mgsin\theta \cdot x |_0^{15} \\<br /> <br /> = 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 :/

 

[Simon Bridge]

why is the distance used to determine torque (which is Force x Distance x sin θ) halfway up the ladder?

Because gravity acts at the center of mass.

 

[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.

<br /> \tau = x \cdot F \cdot \sin\theta \\<br /> F = m \cdot g \\<br /> m = \lambda \Delta x \\<br /> \tau_i = x_i \cdot \lambda \Delta x \cdot g \cdot \sin\theta = \lambda g \sin\theta \cdot x_i \cdot \Delta x \\<br /> \sum_i^n \lambda g \sin\theta \cdot x_i \cdot \Delta x \\<br /> \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\\<br /> = \lambda g \sin\theta \cdot \int_0^{15} x dx = \lambda g \sin\theta \cdot \left[ \dfrac{1}{2}x^2 \right]_0^{15}<br />

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?

 

[Bennigan88]

Because gravity acts at the center of mass.

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

 

[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?"

 

[haruspex]

<br /> \tau = mg \times x \times sin ^\theta = mgsin\theta \cdot x \\<br /> <br /> d\tau = mgsin\theta \cdot dx \\<br />

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):
<br /> d\tau = mgxsin\theta \cdot dx \\<br />

 

[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.
<br /> 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)<br />
... 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 :)

 

[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.