- #1
Bennigan88
- 38
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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:
[tex]
\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[/tex]
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 :/
I tried to use integration to calculate the torque, and this is what I ended up with:
[tex]
\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[/tex]
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 :/
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