A cubic crate of side s=2.0m is top-heavy: its CG is 18cm above its true center.

Part A) How steep an incline can the crate rest on without tipping over?

Part B) What would your answer be if the crate were to slide at constant speed down the plane without tipping over? (The normal force would act at the lowest corner)

To solve both parts of the question, we can use the principle of torque and equilibrium. Since the crate is top-heavy, it is important to consider the center of gravity (CG) and its position relative to the center of the crate.

Part A) To determine the maximum steepness of the incline without tipping over, we need to find the critical angle at which the torque due to gravity is zero. Let's consider the situation when the crate is about to tip over. At this point, the crate is in equilibrium, meaning that the sum of all torques acting on it is zero.

The torque due to the weight of the crate is given by the product of the weight (mg) and the perpendicular distance (d) between the line of action of the weight and the pivot point (CG).

Since the crate is a cube, the CG is located at its center. The distance between the CG and the bottom face of the crate (where the incline will be) is 18 cm or 0.18 m, as given in the question.

The weight of the crate can be calculated as the product of its mass (m) and the acceleration due to gravity (g = 9.8 m/s^2). Since the crate is a solid cube, we can determine its mass using its density.

Let's say the density of the material the crate is made of is ρ. The mass of the crate (m) is then given by the formula: m = ρ * volume.

The volume of a cube can be found by cubing the length of any side, so in this case, the volume is s^3.

Therefore, the mass of the crate is m = ρ * (s^3).

Now, we can calculate the weight (mg) of the crate as:
Weight (mg) = (ρ * (s^3)) * g.

The torque due to the weight can be expressed as:
Torque = (ρ * (s^3)) * g * d.

To prevent the crate from tipping over, this torque must be exactly balanced by the counter-torque produced by the normal force exerted by the incline on the crate. The normal force acts at the lowest corner.

Now, the torque due to the normal force can be calculated using the same formula as above, but with the perpendicular distance (d') between the line of action of the normal force and the pivot point (lowest corner) instead of CG.

For a cube, the distance between any corner and the center of an adjacent face (diagonal of a face) is s * sqrt(2). Therefore, the perpendicular distance (d') is (s * sqrt(2)) - d.

The torque due to the normal force can be expressed as:
Torque_normal = (ρ * (s^3)) * g * [(s * sqrt(2)) - d].

To figure out the maximum steepness of the incline without tipping over, we equate the torques:

(ρ * (s^3)) * g * d = (ρ * (s^3)) * g * [(s * sqrt(2)) - d].

Now, we can solve for d:

d = [(s * sqrt(2)) - d].

This simplifies to:

2d = s * sqrt(2).

Replacing the given values:

2 * 0.18 m = 2.0 m * sqrt(2).

From this equation, we can solve for the steepness of the incline:

Steepness of incline = arctan(d/s) = arctan((0.18 m)/(2.0 m * sqrt(2))).

Calculating this value will give us the maximum steepness of the incline without tipping over.

Part B) If the crate were to slide at a constant speed down the plane without tipping over, the normal force would act at the lowest corner. In this case, the crate experiences no rotational motion, and the torques do not need to be balanced. The frictional force acting up the incline would equal the weight component parallel to the incline.

To calculate the maximum angle at which the crate would slide without tipping, we can use the equation:

Tan(θ) = μ,

Where μ is the coefficient of friction between the crate and the incline. For sliding without tipping, the frictional force would be μ times the normal force acting at the lowest corner.

We can determine the frictional force as:

Frictional force = μ * (ρ * (s^3)) * g.

Comparing the frictional force and the weight component parallel to the incline (mg sinθ), we can set them equal:

μ * (ρ * (s^3)) * g = (ρ * (s^3)) * g * sinθ.

The ρ, s, and g terms cancel out, leaving us with:

μ = sinθ.

From this equation, we can solve for the angle (θ) at which the crate would slide without tipping over.

By calculating the maximum steepness of the incline and the angle for sliding without tipping over, you can determine the answers for both parts of the question.