Question:
Three masses m 1 , m 2 and m 3 are of 100N, 200N and 150N
respectively. The corresponding radii are 0.3 m, 0.15 m and
0.25 m respectively. Angles between masses m 1 and m 2 is
45 o and between m 2 and m 3 is 75 o and between m 3 and m 1
is 240 o . Determine graphically the position and magnitude
of the balance mass required if the radius of rotation is
0.2 m.
Answer:
Given data m 1 = 100N, m 2 = 200 N , m 3 = 150 N , r 1 = 0.3m , r 2 = 0.15 m, r 3 = 0.25m
Radius of rotation = r= 0.2m
Balancing force is equal to resultant force
So, m x r = 63
m x 0.2 = 63
m = 315 N
Measurement θ = 60 0
marks:
ME-6I-22655-W19-Q2-a-5-U6