To solve this problem, we will first consider the forces acting on the rod and then use the conditions of static equilibrium to find the readings on the two spring balances.
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Length and Mass of the Rod:
- Length of the rod, \( L = 80 \text{ cm} \)
- Mass of the rod, \( m = 3.2 \text{ kg} \)
- Weight of the rod, \( W = mg = 3.2 \text{ kg} \times 9.81 \text{ m/s}^2 = 31.392 \text{ N} \)
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Positions of the Balances:
- Location of balance C: \( 20 \text{ cm} \) from one end of the rod
- Location of balance D: \( 30 \text{ cm} \) from the other end, which means it is \( 80 - 30 = 50 \text{ cm} \) from the same end where balance C is located.
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Finding the Center of Gravity: The center of gravity of the uniform rod is located at its midpoint, which is \( 40 \text{ cm} \) from either end.
Now, we can set up the static equilibrium equations. We will denote the readings of balance C and D as \( R_C \) and \( R_D \) respectively.
Step 1: Apply the equilibrium of forces
The sum of the vertical forces must be zero:
\[ R_C + R_D = W \]
Step 2: Apply the moment equilibrium
Taking moments about balance C, we take counterclockwise moments as positive. The moment due to the weight of the rod (acting at its center of gravity) and the moment about balance D must balance:
\[ W \times (50 - 20) = R_D \times (30) \]
Substituting the values we have:
\[ 31.392 , \text{N} \times 30 , \text{cm} = R_D \times 30 , \text{cm} \]
Now we simplify:
- Equations
- From equilibrium of forces: \[ R_C + R_D = 31.392 \quad \text{(1)} \]
- From moments about C: \[ 31.392 \times 30 = R_D \times 30 \quad \Rightarrow \quad 31.392 = R_D \quad \text{(2)} \]
Substitute for \(R_D\) in Eq. (1):
Substituting the value of \(R_D\) from equation (2) into equation (1):
\[ R_C + 31.392 = 31.392 \] Thus: \[ R_C = 31.392 - R_D \] From the moment equation we can plug in the weight of the rod to find \(R_C\) and then substitute \(R_D\):
Solving for \(R_C\) and \(R_D\):
Now we also must relate \(R_C\) and \(R_D\) based on moments around C:
Using \(31.392 , \text{N} \times 30 \text{ cm} = (R_D \times 30\ text{cm})\):
Solve for both readings:
We find;
- \(R_C = \frac{50 \cdot W}{30} = 20.8 , \text{N}\)
- \(R_D = 31.392 - R_C = 10.592 , \text{N}\)
Final Answers:
- Reading on balance C: \(R_C \approx 20.8 \text{ N}\)
- Reading on balance D: \(R_D \approx 10.592 \text{ N}\)
Thus, the readings on the two spring balances are approximately \(20.8 , \text{N}\) for balance C and \(10.59 , \text{N}\) for balance D.