Question

The upward normal force exerted by the floor is 620 N on an elevator passenger who weighs 650 N. What are the reaction forces to these two forces? Is the passenger accelerating? If so, what are the magnitude and direction of the acceleration?

Answer 1

Step-by-step explanation:

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Answer 2

The reaction to the upward normal force is the passengers weight which acts downward oppositely which is 620. The reaction to the passengers weight is the upward normal force. The two force lie in the same axis or along one line so we can take the y-axis which is the total forces should be equal to zero. ΣFy=0

the two forces are acting each other but opposite in direction.

let Fu; is an upward force=620N

Fp; is the downward or passengers weight=650N

ΣFy=Fu+Fp

=(620-650)N

=-30N this shows that the passenger is accelerating downward with the acceleration of

F= ma from the above we can get the mass of the passenger by using this formula. W=mg take g=10 N/kg so 650=10m from this we can divide both sides by 10 to get the mass m. m= 650/10

m=65kg

so using newton's second law F=ma we can find the acceleration of the passenger. Given required solution

F=-30N a=? F=ma a=F/m

m=65kg a=-30N/65kg

a= -0.46 N/kg

The negative sign indicates the acceleration is downward.

Step-by-step explanation: