A 36.8-kg crate rests on a horizontal floor, and a 67.3-kg person is standing on the crate. Determine the magnitude of the normal force that (a) the floor exerts on the crate and (b) the crate exerts on the person.

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To determine the magnitude of the normal force in this scenario, we need to understand that the normal force is the force exerted perpendicular to the surface of contact. In this case, there are two surfaces of contact: the floor and the crate.

(a) The normal force exerted by the floor on the crate:
The normal force exerted by the floor on the crate is equal in magnitude and opposite in direction to the force exerted by the crate on the floor. Since the crate is at rest on the floor, the net force acting in the vertical direction must be zero.
Using Newton's second law (ΣF = ma) and considering only the vertical forces, we have:
ΣF = F_norm - F_gravity = 0
Where F_norm is the magnitude of the normal force exerted by the floor on the crate and F_gravity is the weight of the crate. The weight is given by F_gravity = mg, where m is the mass of the crate and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Therefore:
F_norm - F_gravity = 0
F_norm - mg = 0
F_norm = mg
Substituting the given values, we have:
F_norm = (36.8 kg)(9.8 m/s^2)
F_norm ≈ 360.64 N

So, the magnitude of the normal force exerted by the floor on the crate is approximately 360.64 N.

(b) The normal force exerted by the crate on the person:
The normal force exerted by the crate on the person is equal in magnitude and opposite in direction to the force exerted by the person on the crate. Since there is no vertical acceleration, again the net force acting in the vertical direction must be zero.
Considering only the vertical forces:
ΣF = F_norm - F_person = 0
Where F_norm is the magnitude of the normal force exerted by the crate on the person and F_person is the weight of the person. We can calculate the weight of the person using F_person = mg, where m is the mass of the person (67.3 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Therefore:
F_norm - F_person = 0
F_norm - mg = 0
F_norm = mg
Substituting the given values, we have:
F_norm = (67.3 kg)(9.8 m/s^2)
F_norm ≈ 659.54 N

So, the magnitude of the normal force exerted by the crate on the person is approximately 659.54 N.