A straight rod AB which is 60cm long, is in equilibrium when horizontal and supported at a point C, 10cm from A with masses of 6kg and 1kg attached to the rod at A and B respectively. It is also in equilibrium and horizontal when supported at another pivot at its midpoint, with masses of 2kg and 5kg attached at A and B respectively. Find the mass of the rod and the distance of the centre of gravity from A?
To solve this problem, we can use the principle of moments, which states that for an object in rotational equilibrium, the sum of the anticlockwise moments about any point must equal the sum of the clockwise moments about the same point.
Let's first consider the equilibrium condition when the rod is supported at point C. We can start by finding the total clockwise and anticlockwise moments acting on the rod about point C.
1. Clockwise Moments:
- The 6 kg mass at point A exerts a moment equal to its weight (6 kg x 9.8 m/s^2) multiplied by the perpendicular distance from point A to C, which is 10 cm or 0.1 m. So, the clockwise moment due to the 6 kg mass is (6 kg x 9.8 m/s^2 x 0.1 m).
2. Anticlockwise Moments:
- The 1 kg mass at point B exerts a moment equal to its weight (1 kg x 9.8 m/s^2) multiplied by the perpendicular distance from point B to C, which is the length of the rod minus the distance from A to C (60 cm - 10 cm) or 50 cm, which is 0.5 m. So, the anticlockwise moment due to the 1 kg mass is (1 kg x 9.8 m/s^2 x 0.5 m).
Since the rod is in equilibrium, the total clockwise moments must be equal to the total anticlockwise moments. Therefore, we can set up an equation:
(6 kg x 9.8 m/s^2 x 0.1 m) = (1 kg x 9.8 m/s^2 x 0.5 m)
Now, solve the equation to find the mass of the rod:
(6 kg x 9.8 m/s^2 x 0.1 m) = (1 kg x 9.8 m/s^2 x 0.5 m)
0.588 kg.m = 0.49 kg.m
So, the mass of the rod is 0.49 kg.
To find the distance of the center of gravity (CG) from point A, we can consider the equilibrium condition when the rod is supported at its midpoint.
1. Clockwise Moments:
- The 5 kg mass at point B exerts a moment equal to its weight (5 kg x 9.8 m/s^2) multiplied by the perpendicular distance from point B to the CG. Since the rod is supported at its midpoint, the perpendicular distance is half the length of the rod, which is (60 cm / 2) or 30 cm, which is 0.3 m. So, the clockwise moment due to the 5 kg mass is (5 kg x 9.8 m/s^2 x 0.3 m).
2. Anticlockwise Moments:
- The 2 kg mass at point A exerts a moment equal to its weight (2 kg x 9.8 m/s^2) multiplied by the perpendicular distance from point A to the CG. Since the rod is supported at its midpoint, the perpendicular distance is also half the length of the rod, which is (60 cm / 2) or 30 cm, which is 0.3 m. So, the anticlockwise moment due to the 2 kg mass is (2 kg x 9.8 m/s^2 x 0.3 m).
Again, since the rod is in equilibrium, the total clockwise moments must be equal to the total anticlockwise moments. Therefore, we can set up another equation:
(5 kg x 9.8 m/s^2 x 0.3 m) = (2 kg x 9.8 m/s^2 x 0.3 m)
Now, solve the equation to find the distance of the CG from point A:
(5 kg x 9.8 m/s^2 x 0.3 m) = (2 kg x 9.8 m/s^2 x 0.3 m)
1.47 kg.m = 0.588 kg.m
So, the distance of the center of gravity from point A is 0.588 m.