In a two-dimensional static equilibrium problem, two equations (sum of forces in x and y directions) are needed to solve for the unknowns.

If the shape of an object is altered, the center of gravity may move closer to the heavier side, depending on the distribution of mass.

The acceleration due to gravity near the Earth's surface is approximately 9.81 m/s².

An object moving in a circular path at constant speed experiences centripetal acceleration directed towards the center of the circle.

The angle θ can be found using tan(θ) = μs. Thus, θ = arctan(0.4) ≈ 21.8 degrees.

For a uniform beam, the center of gravity is located at its midpoint, which is L/2.

For a uniform beam, the center of gravity is located at the midpoint of the beam.

For a uniform rectangular plate, the center of gravity is located at the geometric center.

The coefficient of static friction (μs) can be calculated using the formula F = μs * N, where N is the normal force. Here, N = mg = 50 kg * 9.81 m/s² = 490.5 N. Thus, μs = F/N = 200 N / 490.5 N ≈ 0.4.

For a rigid body to be in static equilibrium, both the sum of all forces and the sum of all moments acting on the body must be zero.

An object is in static equilibrium when both the sum of forces and the sum of moments acting on it are zero.

As the angle of inclination increases, the normal force acting on the object decreases.

The frictional force is calculated using the formula F_friction = μN, where μ is the coefficient of friction and N is the normal force.

The moment of a force about a point is calculated as the product of the force and the perpendicular distance from the point to the line of action of the force.

Using Newton's second law, F = ma, we find F = 10 kg * 3 m/s² = 30 N.

Using Newton's second law, F = ma, where m = 5 kg and a = 2 m/s², we find F = 5 kg * 2 m/s² = 10 N.

Friction can help maintain equilibrium by providing the necessary force to counteract any applied forces that would cause motion.

In equilibrium, the sum of all forces acting on an object is zero.

Using the formula h = 0.5 * g * t², we can rearrange to find t = √(2h/g). Substituting h = 45 m and g = 9.81 m/s² gives t ≈ 4.5 s.

Sliding support is not a standard type of support in statics; the common types are fixed, pin, and roller supports.

In the equilibrium of a rigid body, both forces and moments must be considered to ensure the body remains at rest.