Friction and slope

Problem:

A body lays still on a sloping plane with slope angle alpha. How high does the friction coefficient need to be in order to allow for this equlibrium?

Thoughts:

I attempted solving this by basic trig. cos(a) = oposing/hypothenuse. So say the opposing is 1 unit, then the friction coefficient f_k needs to balance the vector acting in the direction of the hypotenuse downwards. So:

cos(a) = 1/f_k

f_k = 1/cos(a).

However the answer is supposed to be f_k = tan(a). How come?

http://ift.tt/1prtzd3

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