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8 Oct 2007 21:24:25 IST

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q 28,29,31&33 of h.c.verma pg 81

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q 28,29,31&33 of h.c.verma pg 81

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waterdemon

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Joined: 11 Jun 2007

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8 Oct 2007 22:25:41 IST

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Q.33)

The Hanging mass is "M" and hence the pulling force is "Mg"

The mass in motion is (m+M+M')

Therefore the acceleration will be:

F= M"a

Where M" = Total Mass

F = (m+M+M')a

Mg = (m+M+M')a

a = Mg/(m+M+M')

The mass "m" will not slide over M' if :

aCos@ = gSin@

Therefore,

MgCos@/(m+M+M') = gSin@

From solving above we get :

M = (m+M')/(Cot@-1)

Hope you find it useful.

plZ do rate me if useful.

Cheers!!!!!!!!!@@@!!!!!!!!!!!

biki ....

Blazing goIITian

Joined: 5 Jun 2007

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8 Oct 2007 22:33:43 IST

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waterdemon has answered 33.
so i think no need of answering it
let me solve 31.

biki ....

Blazing goIITian

Joined: 5 Jun 2007

Posts: 978

8 Oct 2007 23:09:29 IST

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consider the figure shown and observe carefully the forces shown....

for the body of mass M ....

Mg - T = M.2a _________(1)

and as the pulley B is massless .... So force acting on it must be zero....

Such that tension in the string connecting B and 2M is T+T = 2T

So 2T = 2M.a _______(2)

Comparing the two equations ....

Mg - T = 2T

ot T = Mg/3

So (1) => Mg - Mg/3 = 2Ma

or a = g/3

Now accl^n of M = 2a = 2g/3

And force due to clamp on pulley = force on clamp due to pulley... = resultant of the two tensions in the string going over the pulley A

(the string connected to the clamp has no effect due to the pulley on the clamp...but itself pulls the clamp..... and so is not to be counted)

So reqd. force due to pulley on clamp =

( T² + T² )

2T²

= T

= Mg/3 . 2 ( 45⁰ with horizontal as the two forces are equal in magnitude and are mutually perpendicular )

viswesh Nrayanan

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Joined: 9 Oct 2007

Posts: 73

9 Oct 2007 10:37:36 IST

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FOr 28

its easy

T - 1*10 = A (1)
and Tension of the lower sytem is half of the first......Since the pulley is massless T" = T/2

20 -T/2 = 2A -2a .............(2)

30 - T/2 = 3A + 3a............ (3)
Three equations 3 variables solve these..............

u will get A = 10- 5a
Then plug it in other equation
u will get
A = 190/29 .........which is same as 19g/29 upwards
therefore for m1 and m2
17g/29 downwards
21g/29 downwards

Then use s = ut + 1/2 at square

where u is 0 s is 20/100 m solve u will get
20/100 =1/2*9.8* t^2
T=.2459
which can be aprooximated to .25 sec.......

RAte me buddy..................

viswesh Nrayanan

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Joined: 9 Oct 2007

Posts: 73

9 Oct 2007 10:42:25 IST

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FOr 29

USe the equation (2) and (3)

just make it 0

20 -T/2 = 0 As they sud be in rest
30 - T/2 = 0

add you get 50 = T so the mass is 5 kg...........HEy i took g as 10 .....HE took it as 9.8 thats y the difference ..........RATE ME!!!!!!!

biki ....

Blazing goIITian

Joined: 5 Jun 2007

Posts: 978

9 Oct 2007 18:07:43 IST

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we solve for you man.........

but why dont you rate....??????

biki ....

Blazing goIITian

Joined: 5 Jun 2007

Posts: 978

10 Oct 2007 16:56:31 IST

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MR. RAJEEV MASSEY .... you think goiit is an HC Verma solver.........??? a special site where hcv sums are available for free that you give a dozen of questions at a time and people are there to waste time for you ....!!!!!

you are at first violating the norms of this site by giving dozens of questions...... then again you dont give questions....( only numbers )
and when people solve for you..... you show your idiotism by not giving a a single response to it......

anybody might not have any objection to it but I HAVE !!!!!!

and i dont fear boinks..... you want to give me one .... give it...
but i must express my views..!!!!

ramyani chakrabarty

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Joined: 22 Apr 2007

Posts: 2537

10 Oct 2007 23:25:32 IST

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Strong views.
But he gave u 3 hats !!!!!

viswesh Nrayanan

Cool goIITian

Joined: 9 Oct 2007

Posts: 73

14 Oct 2007 20:24:33 IST

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Biki leave it..............

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