 # An elevator starts from rest with a constant upward acceleration.It moves 2m in the forst 0.6 sec.A passanger in the elevator is holding a 3kg package by a vertical string.What is the tension in the string during acceleration.(take g=9.8 m/s^2)

10 years ago

Here find the acceleration of the elevator by using the formula : s=ut + 0.5at2

where u is zero(as it is rest initially) t = 0.6s, s=2m.

then draw the free body diagram for the block.

the weight is acting downwards=mg (initially when the lift is at rest)

since the acceleration of lift is in the upward direction = a

therefore the relative acceleration is a1=(g+a)

then tension in the string T = ma1

Substitute the values to get the answer.

10 years ago

The elevator in this problem is an non inertial frame, and so newtons laws can be directly applied to it.

Pseudo forces should also be taken into consideration.

For the pseudo forces, the acceleration of the block can be found by the following method

= 2 = 1/2(a)(0.6)2

= 4=(0.36)a

=a= (4/0.36)=11.11

Then pseudo forec = 3*(4/0.36)=33.33

thn the tension in the string = 30N(weight of the body)+33.33N(pseudo force)

= 63.33N

10 years ago

@ Swapnil saxena,What is pseudo force?

To all,

I have a doubt,

the acceleration will be acting upwards , the gravitational acceleration and tension will be acting downwards .

Thus it should be mg+t=ma ,na?

How is mg+ma=t ???

10 years ago

Hi Satyaram,

Net Force = Mass*Acceleration.

Hence T - mg = ma.

Net force in the upward Direction = T - mg, which should be mass*acn [Newtons Law].

Assume it is valid.

Regards,

10 years ago

Hi Sathyaram,

Is that so... Just think about it.

Say we have a pulley, with two masses hanging through a rope on either sides of the pulley. Now the heavier mass will move down and the lighter mass will move up.

But still the tension in the rope for both the masses will be in the upward direction.

So it is not required that the tension will be in the opposite direction of motion of the object.

Regards,