Question

A 500 N physics student stands on a bathroom scale in an 850-kg (including the student) elevator that is supported by a cable. As the elevator starts moving, the scale reads 400 N.

(a) Find the acceleration of the elevator (magnitude and direction -- take up to be the positive direction).

(b) What is the acceleration if the scale reads 620 N?

c- If the scale reads zero, what is the acceleration of the elevator?

Should the student worry? Explain your answer.which one is the right?

yes, because the elevator is in free fall

yes, because this great of an acceleration could cause a blackout  

  no, because gravity is absent in his frame

(d) What is the tension in the cable in parts (a) and (c)?

cable tension in part (a)

cable tension in part (c)

Answer 1

the scale reading is the normal force acting on the student

we find the normal force using newton's second law applied to the student:

sum of all forces = ma

the forces acting on the student are the normal force (up) and weight (down), so newton's second law reads:

N - mg = ma or N=m(g+a)

when the elevator is at rest, the normal force equals the weight, so we know that the student's weight is 500N, so that the mass is
500/9.8 =51.02 kg

a) in order for the scale to read 850N, the elevator must be accelerating upward, accoring to newton's second law

N- mg = ma

if N = 850N and mg = 500 N, we have that:

850-500=51.02a or a =6.86 m/s/s

b) for the normal force to be 620N, the elevator must accelerate upward:

620N-500N = ma or a = 2.35 m/s/s

c) if the scale reads zero, then N = 0 and a=-g meaning the elevator is in free fall, and the student should worry

Answer 2

a) without acceleration scale should read 500 N that is weight of student.

now reading is 420 N . so lift is moving downward

acceleration = - (500-420)/50= - 1.6 m/s²

b) reading is greater than weight that is lift is accelerating upward.

acceleration = (620-500)/50 = 2.4 m/s²

c) yes, because the elevator is in free fall.

d) tension in cable , in part a , T = Mg-Ma = 850 * 10 - 850*1.6 = 7140 N

in part c , T = 0. as elevator is in free fall we can assume similar situation where cable breaks that means now tension in cable is zero.

Answer 3

If scale reading is zero, then the acceleration of the elevatoris equal to the acceleration due to gravity thatis10 m/s The tension in the cable in the case of part (a) can be calculated as follows. mgtudent + metevator)g +metevator T=(mtudent +m,levator )(g-a) = (850 kg)(10 m/s, 2 m/s2) - 6800 N The case (c) is the freely falling. So, the tension in the cable is zero.