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