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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.
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
calculate acceleration s=1/2at2
s=2
t=0.6
a=11.1
then drawing fbd
t=ma+mg
t=62.7newton
@ 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 ???
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].
Please dont question Newtons Law.
Assume it is valid.
Regards,
Ashwin (IIT MadraS).
But,always,the tension will be in the direction opp. to direction of motion of object,right?
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.
Ashwin (IIT Madras).
Sorry,I take back my words....
What is pseudo force?
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