XIV
a) If the force (k) works parallel with the inclined plane only
as great a part of the weight (G) is wanted as the height (H) of
the inclined plane is a part of its length (L), hence: K = H/L G.
b) If the force acts parallel to the base (B) of the inclined
plane, K = K/B G. An appli
cation of this plane we see
in the zigzagging of a road
up a hill side 116 and in
the loader. 115.
§ 38. 2) The Wedge. 117. 118. It is a combination of
two inclined planes or it is a straight three-sided prism, the
bases of which are equilateral triangles, the sides of this triangle
we call the sides, and the base the back of the wedge. If the
resistance acts perpendicularly upon the sides of the wedge, K =
½ R/S. G (K= force, G = resistance, R the back and S a side of
the wedge).
Our figure proves this formula. D E re
presents the power, which is decomposed into
the two forces D H and D K, acting vertically
upon the sides A B and C B and contrary
to the resistance; thus, the triangles D E
H and A C B being similar:
D E: D H = A C: A B
DE: 2 DH = ½ A C: A B
D E: 2 D H = A D: A B
K: G = ½ R: S
K = ½ R/S G. — Express this formula
in words.
