In order to achieve the static equilibrium of the see-saw about the fulcrum P, shown in the figure, the weight of the Box B should be ________ kg, if weight of Box A is 50 kg.
To achieve static equilibrium of the see-saw about the fulcrum P, the moments on either side of the fulcrum must be equal. The moment is given by the product of the force and the distance from the fulcrum. For this scenario, the force is represented by the weight of the boxes.
Let's denote:
In equilibrium:
W_A \cdot d_A = W_B \cdot d_B
Substitute the given values:
50 \cdot 5 = W_B \cdot 8
250 = 8W_B
Solve for W_B:
W_B = \frac{250}{8} = 31.25 \text{ kg}
Therefore, the weight of Box B should be 31.25 kg to achieve static equilibrium.

This calculation confirms that the correct answer is 31.25 kg.
A simply supported RCC beam of cross section $0.4 \text{ m} \times 0.6 \text{ m}$ covers a span of $8 \text{ m}$. It is subjected to a uniformly distributed load of $30 \text{ kN/m}$. If the unit weight of concrete is $24 \text{ kN/m}^3$, the tensile stress (in $N/mm^2$, rounded off to two decimal places) at the bottom of the beam at mid-span is______
A rectangular beam section of size 300 mm (width) X 500 mm (depth) is loaded with a shear force of 600 kN. The maximum shear stress on the section in N/mm² is ___________
A simply supported beam AB has a clear span of 7 meter. The bending moment diagram (BMD) of the beam due to a single concentrated load is shown in the figure below.
The magnitude of the concentrated load in kN is __________.