Question

Two friends are carrying a 200-kg crate up a flight of stairs. The crate is 1.25 m long and 0.500 m high, and its center of gravity is at its center. The stairs make a 45.0° angle with respect to the floor. The crate also is carried at a 45.0° angle, so that its bottom side is parallel to the slope of the stairs (Fig. 2). If the force each person applies is vertical, what is the magnitude of each of these forces?

Answer 1

Each person must apply a vertical force of approximately 692.8 N to carry a 200-kg crate up a flight of stairs at an angle of 45°.

Step-by-step explanation:

The student is asking about the force required by each person to carry a 200-kg crate up a flight of stairs at a 45° angle. Both the stairs and the crate are aligned at the same angle, simplifying the situation to a problem of equilibrium where the forces parallel and perpendicular to the incline must balance out.

To find the force each person applies, we consider the weight of the crate and the orientation. The vertical component of the weight (which is the one the carriers are opposing with their force) is the one that must be equaled by the summed forces of both carriers since they are applying a vertical force.

The weight of the crate is mg, where m is the mass of the crate (200 kg) and g is the acceleration due to gravity (9.8 m/s²). The vertical component of this weight is mg sin(θ), with θ being 45°. This gives us the weight component each person must balance:

W_vertical = mg × sin(45°) = (200 kg) × (9.8 m/s²) × sin(45°)

Calculating this, we get W_vertical = 200 × 9.8 × 0.707 ≈ 1385.6 N. The force each person must apply is half of this, as there are two people:

F_person = ½ × W_vertical ≈ ½ × 1385.6 N ≈ 692.8 N

Thus, each person must apply a vertical force of approximately 692.8 N to carry the crate up the stairs.