Solving Elevator Accleration Problems with a Bathroom Scale

[sunshine9489]

a student takes a bathroom scale into an elevator on the 64th floor of a building. The scale reads 836 N.

a. as the elvator moves up the scale reading increases to 935 N, then decreases back to 836 N. Find the accleration of the elevator.

b. As the elevator approaches the 64th floor, the scale reading drops as low as 782 N. What is the acceleratino of the elevator.

help would be greatly appreciated! becuase i feel as if i don't know enough infomation in the problems!

thanks!

 

[vaishakh]

the increase in weight is due to a pseudo force applied by the weight machine suface. the actual force is Mg where m is his mass and g is the accelerationdue to gravity. thus when there is an acceleration of a, then a pseudo force of Ma has to be applied by the machine. be smar enough to understand what i write and convert it into equation. reply back after two days if ou still doesn't get the answer

 

[vaishakh]

the increase in weight is due to the pseudo force applied by the weight machine surface. as a reaction pair normally it applies a normal force of Mg against the person. when lift accelerates, the lift becomes a non-inertial frame due to which the measurementhas error. the magnitude of error is due to pseudo force applied by the peson on the frame against the motion. convert what i write into beneficial equation. tjhere lies your smartness. reply back if you reach nowhere even after two days

 

[daniel_i_l]

Welcome Sunshine. It would also be greatly appreciated if you showed some work!

 

[natural23]

Welcome Sunshine, I am new here also.

Apply F=ma to determine the mass of object on the scale, then use this mass (approx. 85.3 Kg) to determine the total accelerations which include acceleration due to gravity (g). To obtain the elevators acceleration, in (a) and (b), in each case subtract g from the total acceleration :

Mass of object on scale: F=ma -> m = F/a = 836N/9.8m/s^2 = 85.3Kg (approx).

Elevator acceleration in (a): total acceleration = 935N/85.3Kg = 10.96m/s^2 -> acceleration due to elevator = 10.96m/s^2 - 9.8m/s^2 = 1.2m/s^2 (approx).

Elevator acceleration in (b): total acceleration = 782N/85.3Kg = 9.17m/s^2 -> acceleration due to elevator = 9.17m/s^2 - 9.8m/s^2 = -0.63m/s^2 (approx).


David


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