Physics, asked by Anonymous, 10 months ago

✴ Open challenge :
(For all great Phy Lovers !)

→ A person of mass 30kg stands on a weighing scale on a lift. If the lift is descending downwards with a uniform acceleration of a $_1$ , what should be the reading of the weighing scale ?

▪ Mass of lift = 50kg
▪ Force acts on block = 2N
▪ Co-efficient of friction b/w block and surface = 0.5

▶ Don't answer if you don't know !!

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Answers

Answered by Rajshuklakld

Let T1=T/2

so,total tension which is acting at lift=2×T/2=T

from the figure we can clearly say

T2=T/2

T3=T2/2=T/4

T4=2T1=T

since there are six strings which will have same tension T5

so,

6T5=2T3+T4

6T5=T/2 +T=3T/2

T5=T/4

Taking out the tension of single string connectng pulley and block

tension of that string=2T5=T/2

tension of other two strings connected to block=T5/2=T/8

but These three strings which will have tension of T6,so

3T6=T/8 +T/8+T/2=3T/2

3T6=3T/2

T6=T/2

so, tension with which block will be pulled =3T6=3T/2

now,

friction force which will act downward, during movement=2×0.5=1N

3T/2 -(4×10+1)=0.4kg×a2

3T/2 -41=.4a2

3T-82=.8a2....i)

also,

800-T=ma1

2400-3T=3×80×a1

2400-3T=240a1....ii)

now let us find the relation between,,a1 and a2

clearly we can see that L1,L2,L3,L4 will remain constant.....

this can only be possible ,when 6 pulleys close to the lift ,also move with acceleration,,of a1

so,string will move with acceleration a1 upto the last attachment of block,

hence we can say

a1=a2

now, adding eqaution 1) and ii)

we get

2400-82=240.8a1

a1=2318/240.8

a1=9.62

normal or weight that will shown by weighing Machine=mg-ma=m(10-9.6)=.4×30=1.2N

this will be the weight shown by weighing Machine