- #1

- 1

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Is it as simple as using cos or sine of the overall weight of the ladder to find the force on the wall?

i know how to do friction sort of things so we can just assume for this there is no friction..

any help is appreciated, thanks :)

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- Thread starter big_man4141
- Start date

- #1

- 1

- 0

Is it as simple as using cos or sine of the overall weight of the ladder to find the force on the wall?

i know how to do friction sort of things so we can just assume for this there is no friction..

any help is appreciated, thanks :)

- #2

- 24

- 0

Also, consider the force of gravity to be acting on the center-of-mass of the ladder, and we have:

[tex]Torque = \textbf{F}\times\textbf{R}[/tex]

[tex]\tau = <0, -mg>\times<d/2, \frac{\sqrt{L^2 - d^2}}{2}>[/tex]

[tex]\tau = mgd/2[/tex]

[tex]\tau = <0, -mg>\times<d/2, \frac{\sqrt{L^2 - d^2}}{2}>[/tex]

[tex]\tau = mgd/2[/tex]

Assuming you're familiar with that vector notation. It's pretty straight-forward.

-Jake

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