Question

A pendulum bob of mass 50 g is suspended from the ceiling of an elevator. Find the tension in the string if the elevator (a) goes up with acceleration 1.2 m/s2, (b) goes up with deceleration 1.2 m/s2, (c) goes up with uniform velocity, (d) goes down with acceleration 1.2 m/s2, (e) goes down with deceleration 1.2 m/s2 and (f) goes down with uniform velocity.

Answer 1

M = 50 gm,    g = 10 m/s^2

acceleration downwards of the elevator ,  a = 1.5 m/s^2  

Frame of reference:

      non-inertial frame.  A coordinate access attached to the elevator.

  

   In the non-inertial frame, we add a pseudo force on each object having a mass in the direction opposite to the movement of the frame.  So add a pseudo force on pendulum upwards = m a = 0.050 * 1.5 = 0.075 N

There is the weight of pendulum m g = 0.50  N  acting downwards.

There is tension T in the string acting upwards on the pendulum.

As the pendulum is at rest in the non-inertial frame:

      T + 0.075 N = 0.50 N

      T = 0.425 N

Answer 2

Explanation:

(a) When the elevator goes up with acceleration 1.2 m/s²

T=mg+ma

⇒ T = 0.05 (9.8 + 1.2)

= 0.55 N

(b) goes up with deceleration 1.2 m/s2:

T=mg+m(-a)=m(g-a)

⇒ T  = 0.05 (9.8 − 1.2)

= 0.43 N

(c) goes up with uniform velocity:

T=mg⇒ T = 0.05 × 9.8 = 0.49 N

d) goes down with acceleration 1.2 m/s2:

T+ma=mg

⇒T=m(g-a)

⇒ T = 0.05 (9.8 − 1.2)

= 0.43 N

f)Goes down with uniform velocity:

T=mg

⇒ T = 0.05 × 9.8 = 0.49 N