Math Snap

STEP 1

1. The beam is uniform and has a mass distribution of $50 \, \text{kg/m}$.
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 $9.81 \, \text{m/s}^2$.

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

STEP 4

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

Apply the equilibrium conditions to solve for the reactions at the supports:
- Let $R_A$ and $R_B$ be the reactions at supports A and B, respectively.
- Sum of vertical forces: $R_A + R_B = 1814.85 \, \text{N} + 2943 \, \text{N}$
- Sum of vertical forces: $R_A + R_B = 4757.85 \, \text{N}$
- 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 $R_B$ 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:
$$R_A = 1940.39 \, \text{N}$$ $$R_B = 2817.46 \, \text{N}$$