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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: HC Verma Pg no. 79 Newton's laws of motion Exercise Q no.12
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coolboy910 (5)

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HC Verma Pg no. 79 Newton's laws of motion Exercise Q no.12
    
varun.tinkle (1069)

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IT IS SIMPLE
SINCE THE FORCE MAKES AN @ WITH THE VERTICAL
2FCOS@=MG
F=MG/2COS@
AS HE IS GETTING PULLED @ INCREASES COS @ DECREASES THUS THE FORCE INCREASES
IN THE 2ND CASE
COS@=h/UNDER ROOT H^2+D^2/4
SOLVE IT AND THE ANSWER
WILL COME AS
GIVEN IN THE BACK
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debmalya.roychoudhuri (70)

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hiii frnds please couls u solve hc verma page 82 no 34 and no 33 (part 1)newtons laws i would begrateful to u
rates assured
thank u

there is no way to iit rather iit is the way !!
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plzzz solve mah problems folks i need the solutions i have wracked mah brains for a long time on these two plzzz.....

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pransmuks12 (0)

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Re:HC Verma Pg no. 79 Newton's laws of motion Exercise Q no.12
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varun.tinkle (1069)

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34)

a)

SEEING FROM THE FIGURE

WE CAN SAY THAT ACC OF THE 4 KG BLOCK IS 2 TIMES MORE THAN THAT OF 5KG ONE


LET THE ACC OF 5 KG BLOCK IS A THEN THAT OF THE 4 KG BLOCK IS 2A

50-2T=5A (2T IS THERE SINCE THE BLOCK IS CONNECTED BY THE STRINGS)

T-40=4(2A)

T-40=8A

2T-80=16A

SOLVING IT WE GET

-30=21 A

A= -10/7= -G/7

THIS MEANS THAT THE 5KG BLOCK IS ACC UPWARDS



AND THE 4 KG DOWNWARDS


THE ACC OF 5 KG BOCK IS G/7 AND THAT OF THE 4 KG BLOCK IS 2G/7



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varun.tinkle (1069)

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34)



B)



SEEN FROM THE FIGURE THE ACC OF 2KG BLOCK IS TWICE THAT OF THE 5KG BLOCK


 


LET THE ACC OF THE 5 KG BLOCK BE A THEN THAT OF 2KG BLOCK IS 2A



5G-2T=5A   (2T SINCE IT IS CONNECTED BY 2 STRINGS)



T=2(2A)



T=4A



5G=13A



A=5G/13



THIS IS THE ACC OF THE 5 KG BLOCK



THE ACC OF 2 KG BLOCK IS 10/13 G



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From shadows a light shall spring
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The crown less again shall be king.

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varun.tinkle (1069)

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34)
C)
SEEING FROM THE FIGURE WE CAN SAY THAT THE ACC OF THE 2 KG BLOCK IS TWICE OF THAT OF THE 1KG BLOCK

IF THE ACC OF THE 1 KG BLOCK IS A THEN THAT OF THE 2KG IS 2A

2T-G=A (SINCE IT IS CONNECTED BY TWO STRINGS)
2G-T=2(2A)
2G-T=4A
4G-2T=8A
SOLVING WE GET
3G=9A
A=G/3
THIS IS THE ACC OF THE 1 KG BLOCK THEN THT OF THE 2KG BLOCK IS 2A
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The crown less again shall be king.

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varun.tinkle (1069)

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SEE FROM THE FBD OF THE BLOCK m
U CAN SEE THAT
MACOS@=MGSIN@
A=GTAN@
SINCE BOTH THE MASS mAND MASS M" ARE ACC TOGETHER THEY CAN BE CONSIDERED AS THE SAME SYSTEM
SO THIS MUST ALSO BE THE ACC OF THE BLOCK M
MG-T=MGTAN@
MG-MGTAN@=T
MG(1-TAN@)=T
T=(M'+m)TAN@
SOLVE IT AND GET THE ANSWER
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The crown less again shall be king.

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varun.tinkle (1069)

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ONE MORE METHOD TO SOLVE THE 33 SUM

use this in Mg - T = Ma
or M = T/(g - a) to get the answer....
T - nsin = M'a
=>T = M'a + nsin
=> T = M'a + (masin + mgcos).sin
= M'g.tan + mg.tan.sin2 + mg.cos.sin ..... using a = g.tan
= M'g.tan + mg (tan.sin2 + cos.sin)
= M'g.tan + mg (sin3/cos + cos.sin)
= M'g.tan + mg ( sin3 + cos2.sin ) / cos
= M'g.tan + mg (sin/cos).(sin2 + cos2)
= M'gtan + mg.tan
= (M'+m).g.tan

use this in Mg - T = Ma
or M = T/(g - a) to get the answer....
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From shadows a light shall spring
Renewed shall be blade that's broken
The crown less again shall be king.

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magiclko (4200)

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i think ur problems r solved...

 






but do try to post the questions here, so dat the gang can solve it for u..i mean even the members who dnt hav the particular buk

 





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coolboy910 (5)

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for sum no.33 can any1 xplain hw macos = mgsin cum.......................


an FBD diagram will hepl me

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