# Elvator tension problem

1. Feb 26, 2005

### runner1738

elevator tension problem

An elevator starts from rest with a constant upward acceleration and moves 1 m in teh first 1.5s. A passenger in the elevator is holding a 9.8 kg is holding a 9.8 kg bundle at the end of a vertical cord. G=9.8 m/s^2. What is the tension in the cord as teh elevator accelerates? Answers in units of N.

my work: (9.8kg * g)((4/9 m/s^2)/g + 1) = 96.04(1.045351474)=100.39555556 the 4/9 is the acceleration in the y direction 1/1.5/1.5 so m/s^2 right? this wasnt right what is wrong?

Last edited: Feb 26, 2005
2. Feb 26, 2005

### runner1738

t=MAy+ mg? so im guessing the thing i have wrong is acceleration right?

3. Feb 26, 2005

### runner1738

the answer was 104.751 someone said a=v^2/2d then use a in ma+mg but that never gave me 104.751 unless i just cant add, anyone?

4. Feb 27, 2005

well first you need to find the acceleration of the elevator

xf=xi + vi + 1/2 at2

1m=0 + 0 + .5a(1.5)2 so a= 1m/(.5 x 2.25s2)

this will be vectorally added with the acceleration due to gravity
then tension will be equal to the weight T=weight=m(g+a)

T=9.8kg x (9.8 m/s2 + 1m/.5 x 2.25s2)

that'll give you exactly 104.7511111 N

just remember in these cases, the elevators acceleration just adds to gravity
if the elevator was heading down you need to subtract it

Last edited: Feb 27, 2005