IIT JEE , AIEEE Forum - Physics , Mechanics

asish

any one having HCV part1 can u pls do problem no. 21&28 from the chapter friction
page no.-98&99.thought of typing the question but they have figures which i cant draw here so please do them and give soln

anchitsaini

for man--
F+N=(M+m)g

for block-
F=f=uN

from above--
F+F/u=M+m)g
therefore
F=u(M+m)g/(u+1)

anchitsaini

i'll post 28 tomorrow
its a bit long

smartboy198062

in Q 21

let the man pull with a force T ................ due to this force he exerts on the rope he reduces the normal reaction by T.

Then for the stage where tension is just able to pull the slab

u((M+m)g-T)=T

u(M+m)g=T(1+u)

T=u(M+m)g/(1+u) (ans)

Greatdreams

Well dude its my discredit that I cannot provide the diagram but I guess you know how to draw the free body diagrams....

So first draw the free body diagrams

Now a = accn of the block M in horiz dir

R₁ = normal reacn between the 2 blocks

R₂ = normal reacn btw the bigger block and the ground

T = Tension in the string

So from the free body diagram of the smaller block,

we have R₁ = ma

& mg - T -

₁ R₁ = 2ma

So T = mg -

₁R₁ - 2ma.............................(1)

Now come to the free body diagram of the bigger block

So you have

R₂ - Mg -

₁R₁ = T.........................................(2)

Now putting the value of T and R₁ we have

R₂ = Mg +

₁ma + mg -

1R₁ - 2ma

So R₂ = Mg + mg - 2ma...........................................(3)

Again from free body diagram of the bigger block,

we have,

2T - R₁ -

₂R₂ = Ma

So now putting the values of R1 and R2

we have

2T - ma -

₂(Mg-mg-2ma) = Ma

so 2T = (m+M)a +

₂(Mg+mg-2ma) ..............................(4)

Now considering eqn (1) and (4)

(m+M)a +

₂(Mg+mg-2ma) = 2[ mg -

₁R₁ - 2ma]

So (m+M)a +

₂(Mg+mg-2ma) = 2mg - 2

₁R₁ - 4ma

So 2mg -

₂(M+m)g = a[M+m+4m-2

₂m + 2

₁m]

So we have ,

a = [ 2m -

₂(M+m) ]g / [ M+m(5+2(

₁ -

₂)]

asish

but prob is i am not able to get reln betn accln's of two blocks can u pls tell about that

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