physics

the roof on a house rises 1 .00 m over a horizontal distance of 3.5 m. a 71 kg roofer stands on the roof. is the frictional force that keeps the roofer from slipping equal in magnitude to fparallel or fperpendicular? what is the magnitude of this force?

I'm pretty sure its fparallel, but how do you find fparallel when it gives you a distance? the formula i've been using is fparallel= mg sin(theta)

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  1. friction has to be parallel to the roof.

    Friction=mu*mg*cosTheta

    where theta= arctan(1/3.5)
    solve for theta, then cos theta. If you wish to avoid trig, if tantheta=1/3.5, then cosine=3.5/sqrt(1+3.5^2) check my triangle on that.

    Dont forget about the portion of weight sliding down the plane... mgSinTheta

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    bobpursley
  2. pinecone

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