Quick Reply

Blazing goIITian

Joined: 14 Nov 2007

Posts: 309

20 Mar 2008 00:22:35 IST

Like 3 people liked this

33)

first consider the free body diagrams
now for the mass M,
from fbd, T = Mg - Ma
For the combined mass (m+M')
T = (M'+m)a
equating both Mg =(M' + M + m)a
so a = Mg / (M' + M + m)

now consider the fbd of the mass m ,
ma cos x = mg sin x
so a = g tan x
Now equating this a with the formerly obtained a,
Mg/(m+M+M') = g tan x
so M cot x = M + M' +m

M = (M'+m/ cot x - 1)
ans

Blazing goIITian

Joined: 10 Sep 2007

Posts: 1319

20 Mar 2008 00:22:47 IST

Like 2 people liked this

sum no 25

considre d bloc on d 3m side asA

n on 4 m side asB

thus

"A"

mgsin53-T=ma

"B"

T-mgsin37=ma

thus solvin thenm u get

a=g/10m/s^@

hope u finda usefu!!

Blazing goIITian

Joined: 10 Sep 2007

Posts: 1319

20 Mar 2008 00:27:54 IST

Like 2 people liked this

35th

let dT betwn 50g n 100g b T1

n 100g n 500 g bT2

"50g"

m1g-T1=m1a

"100G"

T1+m2gsin@-T2=m2a

"500g"

T2-m3g=m3a

thus solvin these 3 equations

u'll egt d ansewr

i.e a=400g/650

=8g/13

Cool goIITian

Joined: 10 Feb 2008

Posts: 57

20 Mar 2008 00:37:11 IST

Like 0 people liked this

@Balganesh

How did u calculated ma cos x for fbd of m in q no 33?? i did it as ma. pls tell.

@deedee

how did calculated the angle 53 & 37?? i was stuck in that.

Blazing goIITian

Joined: 14 Nov 2007

Posts: 309

20 Mar 2008 00:38:39 IST

Like 1 people liked this

forM

Mgsin - T =Ma1 ...........1

for2M

2T - 2Mg =2Ma2 .............2

but a1 =2a2 ......3

subst 3 in 2 and adding we het a1 = -g/3

which means the mass M goes up the plane

please rate me for my efforts

Blazing goIITian

Joined: 10 Sep 2007

Posts: 1319

20 Mar 2008 00:39:10 IST

Like 1 people liked this

we get a triangel

wer d sides r 3,4,5 n d ansgle r 37de ,53 de,90de

c d angel bewtn 3 m n 5m is 53de

d angle betwn 4 m n 5 m is 37de

n itz a right angled triangel

hope u get it

try 2 drw it!!

__________________

Blazing goIITian

Joined: 14 Nov 2007

Posts: 309

20 Mar 2008 00:46:03 IST

Like 1 people liked this

since the mass M is attached to the system it will move with accln a in horizantal directn and even m will move with the same accln but since it is not moving down , there should be balancing of all the forces in that directn

waterdemon

Forum Expert

Joined: 11 Jun 2007

Posts: 1048

20 Mar 2008 00:53:26 IST

Like 2 people liked this

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!!!!!!!!!@@@!!!!!!!!!!!

waterdemon

Forum Expert

Joined: 11 Jun 2007

Posts: 1048

20 Mar 2008 00:54:09 IST

Like 2 people liked this

31)

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

= T2

= 2Mg/3

(45⁰ with horizontal as the two forces are equal in magnitude and are mutually perpendicular )

diagram

diagram

Cool goIITian

Joined: 10 Feb 2008

Posts: 57

20 Mar 2008 00:57:21 IST

Like 0 people liked this

thanx both of u. pls help me with the rest of the questions as well.

Blazing goIITian

Joined: 14 Nov 2007

Posts: 309

20 Mar 2008 00:57:35 IST

Like 0 people liked this

hey anish for the 31) u want all three sub questions or only c) reply me fast so that i will edit my this post and give the soln

waterdemon

Forum Expert

Joined: 11 Jun 2007

Posts: 1048

20 Mar 2008 01:03:45 IST

Like 1 people liked this

Q.28)

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(at²)

where u = 0

S = (20/100)m

Now Solve and then you will get
20/100 =(1/2)(9.8)(t²)
T 0.25 sec

Hope it is useful.

Cheers !!!!!!!!!!!!!!!!!

Blazing goIITian

Joined: 10 Sep 2007

Posts: 1319

20 Mar 2008 01:04:40 IST

Like 0 people liked this

here's d sol for 31 (C) part

F net=(T1^@ +T2^@)^1/2

so T1=Mg/3

n T2=Mg/3

so substitutin it theer u get

Fnet=(2)^1/2 Mg/3

itz at an angle 45 coz

wen d 2 forces of tension will act they 'll hav theri resultant in third quadrant at an angle 45 de

Cool goIITian

Joined: 10 Feb 2008

Posts: 57

20 Mar 2008 01:07:52 IST

Like 0 people liked this

@ Balganesh. i want all 3 of them.

Blazing goIITian

Joined: 14 Nov 2007

Posts: 309

20 Mar 2008 01:08:23 IST

Like 1 people liked this

watedemon has posted the soln to 31) all and 34 a) so posting

34 c

for 2kg

2g -T =2a1.........`1

and for 1kg -g +T =a2 .........2

but a1= 2a2.......3

from these eqns we get a1 =2g/3

and a2 = g/3

Cool goIITian

Joined: 10 Feb 2008

Posts: 57

20 Mar 2008 01:15:31 IST

Like 0 people liked this

@balganesh. how did u got a1=2a2 ???

Blazing goIITian

Joined: 14 Nov 2007

Posts: 309

20 Mar 2008 01:16:02 IST

Like 1 people liked this

34 b

for 5kg 5g - T = 5a1.........1

for 2kg 2a2 =T ...........2

and 2a1 =a2........3

from these eqns we get a1 =5g/13

anda2 =10g/13

phew its over i am tired ,a slow typer

Blazing goIITian

Joined: 10 Sep 2007

Posts: 1319

20 Mar 2008 01:18:56 IST

Like 2 people liked this

arey

c 34 c)

hw oen gets a1=2a2

c d string attached 2 1kg

if it goes x up wards d string attached 2 2kg

goes 2x downward thus

s1=2s2

similarly

a1=2a2