A uniform metal rod of length 80cm and mass 3.2kg is supported by horizontally by the two vertical springs balance C and I D.balance C is 20cm from one end while balance D is 30cm from the other end find the reading on each balance
C+D = 3.2
20 C + 50 D = 40 * 3.2
20 (3.2-D) + 50 D = 128
30 D = 64
D =2.13
C = 3.2 - 2.13 = 1.07
To find the reading on each balance, we need to consider the moments acting on the rod.
Let's assume that the distance from the left end of the rod to the center of mass is x.
Therefore, the distance from the right end of the rod to the center of mass is (80 - x).
1. Moment due to balance C (clockwise):
The weight of the rod acts at its center of mass. So, the weight can be considered to act downwards at the center of mass, which is x cm from balance C.
The moment due to balance C can be calculated as:
Moment_C = weight * distance = mg * x
2. Moment due to balance D (anticlockwise):
Similarly, the weight of the rod can be considered to act downwards at the center of mass, which is (80 - x) cm from balance D.
The moment due to balance D can be calculated as:
Moment_D = weight * distance = mg * (80 - x)
Since the rod is in equilibrium, the moments due to balance C and balance D must be equal:
Moment_C = Moment_D
Setting up the equation:
mg * x = mg * (80 - x)
Cancelling the common factors:
x = 80 - x
Simplifying:
2x = 80
x = 40 cm
Now that we have found the position of the center of mass, we can determine the readings on each balance.
Reading on balance C = weight of the left half of the rod = (mass of left half) * g
Reading on balance C = (3.2 kg / 2) * 9.8 m/s^2 = 15.68 N
Reading on balance D = weight of the right half of the rod = (mass of right half) * g
Reading on balance D = (3.2 kg / 2) * 9.8 m/s^2 = 15.68 N
Therefore, the reading on balance C is 15.68 N, and the reading on balance D is also 15.68 N.
To find the reading on each balance, we can analyze the torques acting on the metal rod.
Let's assume that the metal rod is in equilibrium, meaning that it is neither rotating nor accelerating. In this case, the sum of the torques acting on the rod must be zero.
The torques on the rod are created by the gravitational force acting on the center of mass of the rod and the forces exerted by the two balances (C and D).
Let's denote the distance between the center of mass of the rod and balance C as x, and the distance between the center of mass and balance D as (80cm - x).
The torques created by the forces at balance C and balance D can be calculated by multiplying the respective forces by their perpendicular distances from the center of mass.
According to the principle of torques, the torques created by balance C and balance D must balance each other out. Therefore, we have the equation:
Torque by balance C = Torque by balance D
F_C * x = F_D * (80cm - x)
Given that balance C is 20cm from one end, we can substitute x = 20cm into the equation:
F_C * 20cm = F_D * (80cm - 20cm)
Now we can solve for the reading on each balance by rearranging the equation:
F_C = (F_D * 60cm) / 20cm
Additionally, we know that the sum of the readings on both balances must equal the weight of the rod, which is given by:
Weight = mass * gravity
Weight = 3.2kg * 9.8m/s^2
Finally, we can set up another equation using the above information:
F_C + F_D = Weight
Now we have two equations with two unknowns (F_C and F_D) that we can solve simultaneously to find the readings on each balance.