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Math  /  Geometry

Question3/43 / 4 The uniform beam has a mass of 50 kg per meter of length. Determine the reactions at the supports.
Problem 3/43 / 4

Studdy Solution

STEP 1

1. The beam is uniform and has a mass distribution of 50 kg/m 50 \, \text{kg/m} .
2. The beam is supported at two points, A and B.
3. A 300 kg crate is placed on the beam.
4. The acceleration due to gravity is 9.81 m/s2 9.81 \, \text{m/s}^2 .

STEP 2

1. Calculate the total weight of the beam.
2. Calculate the weight of the crate.
3. Apply the equilibrium conditions to solve for the reactions at the supports.

STEP 3

Calculate the total weight of the beam:
- Length of the beam: 3.7 m 3.7 \, \text{m} - Mass per meter: 50 kg/m 50 \, \text{kg/m} - Total mass of the beam: 50 kg/m×3.7 m=185 kg 50 \, \text{kg/m} \times 3.7 \, \text{m} = 185 \, \text{kg} - Total weight of the beam: 185 kg×9.81 m/s2=1814.85 N 185 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 1814.85 \, \text{N}

STEP 4

Calculate the weight of the crate:
- Mass of the crate: 300 kg 300 \, \text{kg} - Weight of the crate: 300 kg×9.81 m/s2=2943 N 300 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 2943 \, \text{N}

STEP 5

Apply the equilibrium conditions to solve for the reactions at the supports:
- Let RA R_A and RB R_B be the reactions at supports A and B, respectively. - Sum of vertical forces: RA+RB=1814.85 N+2943 N R_A + R_B = 1814.85 \, \text{N} + 2943 \, \text{N} - Sum of vertical forces: RA+RB=4757.85 N R_A + R_B = 4757.85 \, \text{N}
- Taking moments about point A: R_B \times 3.7 \, \text{m} = 1814.85 \, \text{N} \times \frac{3.7}{2} \, \text{m} + 2943 \, \text{N} \times 2.4 \, \text{m} \] R_B \times 3.7 = 3357.395 + 7063.2 \] R_B \times 3.7 = 10420.595 \] R_B = \frac{10420.595}{3.7} = 2817.46 \, \text{N} \]
- Substitute RB R_B back into the sum of vertical forces: R_A + 2817.46 = 4757.85 \] R_A = 4757.85 - 2817.46 = 1940.39 \, \text{N} \]
The reactions at the supports are:
RA=1940.39 N R_A = 1940.39 \, \text{N} RB=2817.46 N R_B = 2817.46 \, \text{N}

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