To find the reactions at supports \( R_1 \) (at point C) and \( R_2 \) (at point D) for the plank and the load system, we can use the principles of static equilibrium, specifically:
- The sum of vertical forces must be zero.
- The sum of moments about any point must be zero.
Given:
- Length of plank (\( L \)) = 3.0 m
- Weight of plank (\( W_p \)) = 20 kg = \( 20 \times 9.81 = 196.2 \text{ N} \)
- Weight of load (\( W_l \)) = 10 kg = \( 10 \times 9.81 = 98.1 \text{ N} \)
- Distance from A to center of gravity of plank = 2.0 m
- Distance from A to point C (left support) = 0 m
- Distance from A to point D (right support) = 3.0 m
Steps to Calculate \( R_1 \) and \( R_2 \):
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Set Up the Force Equation: \[ R_1 + R_2 = W_p + W_l \] \[ R_1 + R_2 = 196.2 + 98.1 = 294.3 \text{ N} \tag{1} \]
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Set Up the Moment Equation: Taking moments about point C (to eliminate \( R_1 \)): \[ \sum M_C = 0 \] The moments caused by the weights should equal the moment caused by \( R_2 \): \[ W_p \text{ moment at C} + W_l \text{ moment at C} - R_2 \text{ distance to D} = 0 \]
- Moment due to weight of the plank about C (since its CG is at 2.0 m from A and 1.0 m from C): \[ W_p \times 1.0 \text{ m} = 196.2 \times 1.0 = 196.2 \text{ Nm} \]
- Moment due to weight of the load about C (this is at A, which is 3.0 m from C): \[ W_l \times 3.0 \text{ m} = 98.1 \times 3.0 = 294.3 \text{ Nm} \]
- Moment due to \( R_2 \) (acting at a distance of 3.0 m): \[ R_2 \times 3.0 \text{ m} \]
Setting up the moments equation: \[ 196.2 + 294.3 - R_2 \times 3.0 = 0 \] \[ 490.5 = R_2 \times 3.0 \] \[ R_2 = \frac{490.5}{3.0} = 163.5 \text{ N} \tag{2} \]
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Substituting Back to Find \( R_1 \): Using equation (1): \[ R_1 + 163.5 = 294.3 \] \[ R_1 = 294.3 - 163.5 = 130.8 \text{ N} \tag{3} \]
Results:
Thus, the reactions at the supports are:
- \( R_1 = 130.8 \text{ N} \)
- \( R_2 = 163.5 \text{ N} \)