The position of center of mass of three masses 2 kg, 3 kg and 15 kg placed with respect to mid point (p) of normal bisector, as shown in the figure is . [figure]
- ( 34, 1.25 )
- ( 34, 1.0 )
- (0, 0)
- (1.25, 0)
Answer
(1) ( 34, 1.25 )
Solution
[figure]
Each half of the 120° angle is 60°:
Half base =10 60 =5 3 m, height =10 60 =5 m
P is the mid-point of the altitude ⇒ 2.5 m above the base.
Origin at P: 2 kg at (-5 3, -2.5), 3 kg at (5 3, -2.5), 15 kg at (0, 2.5); total mass 20 kg
xcm= 2(-5 3)+3(5 3)+15(0)20= 5 320= 34
ycm= 2(-2.5)+3(-2.5)+15(2.5)20= -5-7.5+37.520= 2520=1.25
∴ Centre of mass = ( 34, 1.25 )
Answer: (1)