3 forces X,Y,Zacting at the vertices A,B,Cres.of a triangle,each perpendicular to the opp.side,keep it in equi

Question

Three forces X,Y,Z acting at the vetices A,B,C respectively of a triangle,each perpendicular to the opposite side,keep it in equilibrium.Prove that X\a=Y\b=Z\c.(statics)Lesson-Forces Acting At A Point

Answer 1

Let v1=C-B , v2=A-C , v3=B-A be the three displacement vectors making up the sides of the triangle. Define f1,f2,f3 to be the three forces X,Y,Z each rotated 90 degrees counterclockwise. Then f1 is parallel to v1, and so on. That is:

f1 = s1*v1
f2 = s2*v2
f3 = s3*v3

where s1,s2,s3 are three scalars. It's only necessary to show that s1=s2=s3 to prove that the three forces are proportional to the three opposite side lengths (|f1|/|v1|=|f2|/|v2|=|f3|/|v3|). This is easily done as follows:

Since the forces are in equilibrium (f1 + f2 + f3 = 0):

s1*v1 + s2*v2 + s3*v3 = 0

Multiplying (v1+v2+v3)=0 by s1 and subtracting from the above equation:

(s2-s1)*v2+(s3-s1)*v3 = 0

Since v2 and v3 are linearly independent both coefficients must be zero

-->

s1 = s2 = s3
QED.

Answer 2

Since they are in equilibrium they must meet in a point. That point is inside the triangle. Now, draw the diagram and label the angles between the force vectors. Derive reltions with the angles and the sides using Sine's Law for triangles and relate the equations.