Question

A cubic crate of side s = 2.3 m is top-heavy: its CG is 20 cm above its true center.

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

Theta?

2) What would your answer be if the crate were to slide at constant speed down the plane without tipping over? [Hint: The normal force would act at the lowest corner.]

Theta?

Answer 1

To find the maximum angle the crate can rest on an incline without tipping over, calculate the angle at which the vertical line through the CG intersects the side of the crate.

Step-by-step explanation:

To find the maximum angle theta at which the crate can rest on an incline without tipping over, we need to consider the stability of the crate. The center of gravity (CG) of the crate is 20 cm above its true center, which means that the crate will tip over if the vertical line through the CG falls outside the base of the crate.

The base of the crate can be thought of as a rectangular cross-section with dimensions s × s, where s is the side length of the cube. The maximum angle at which the crate can rest without tipping over is the angle at which the vertical line through the CG intersects the side of the crate.

This angle can be found using the tangent function: sin(theta) = CG/base, where CG is the distance of the CG above the true center and base is the side length of the crate. Given that CG is 20 cm and base is 2.3 m, we can calculate the value of theta.

Answer 2

To determine how steep an incline the crate can rest on without tipping over, calculate the torque due to the weight and the torque due to the CG being above the true center. To determine the angle at which the crate will slide at a constant velocity, use the concept of static equilibrium.

Step-by-step explanation:

To determine how steep an incline the crate can rest on without tipping over, we need to consider the torque exerted by the weight of the crate and the torque exerted by the center of gravity (CG) being above the true center. The torque due to the weight can be calculated by multiplying the weight of the crate by the perpendicular distance from the CG to the edge of the base. The torque due to the CG being above the true center can be calculated by multiplying the weight of the crate by the perpendicular distance from the true center to the CG. The crate will tip over when the torque due to the CG is greater than or equal to the torque due to the weight. Using the given dimensions and assuming the crate is symmetrical, we can calculate the maximum angle (theta) at which the crate can rest without tipping over.

For the second question, where the crate slides down the incline at constant speed without tipping over, the normal force on the crate will act at the lowest corner. We can use the concept of static equilibrium to determine the angle (theta) at which the crate will slide at a constant velocity.