A36.9-kg crate rests on a horizontal floor, and a 64.9-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.

FN = 997.64 N :Force that the floor exerts on the crate

Fcp = 636.02N, vertical and upaward

Explanation:

Wp: person Weigth (N)

Wc: crate Weigth ( N)

FN: Normal force

Calculated of the weight

Wc=(36.9kg)(9.8 m/s²) = 361.62 N

Wp= (64.9kg)(9.8 m/s²) = 636.02N

(a) Force that the floor exerts on the crate

For free body diagram of the crate+person

∑Fy=0

FN-Wc-Wp = 0

FN-361.62 N-636.02N= 0

FN = 997.64 N :Force that the floor exerts on the crate

(b)  For free body diagram of the crate

Fpc: Force of the person on the box

∑Fy=0

FN-Wc-Fpc = 0

997.64 N-361.62 N = Fpc

Fpc =636.02N :Force of the person on the crate(vertical and downaward)

Newton's third law or principle of action and reaction :

Force of the person on the crate = -  Force of the crate on the person

Fpc = - Fcp

Fcp:Force of the crate on the person

Fcp = 636.02N, vertical and upaward

i don't understand. explain it clearly