Newton's Laws and WPE (Doubts)

pata nahi answer wrong kyu ara hai LOL [3]
btw euclid i tried wid d same way dat u r telling , but pata nai ans match nai hora

btw vivek wats d source of the question ?

i solved it dis way
a_B/C= [- a_B/C cosθ] (i ) + [- a_B/Csinθ] (j)

(since B moves only along d incline as seen by C )

a_C = a_C (i )

→ a_B = a_B/C + a_C = [a_C - a_B/C cosθ] (i ) + [- a_B/Csinθ] (j)

a_A = a_A (j ) [ no i component as no force in x direction )

also , since A never loses contact with B, A has accn only in horizontal direction AS SEEN BY B ,hence a(_A/B)_y= 0 i.e (a_B)_y = (a_A)_y

so a_A = [- a_B/Csinθ] (j)

now Σ m a _x-component= 0

so a_C+ a_C - a_B/C cosθ = 0

a_B/C cosθ = 2a_C

so a_B = - a_C (i )+ [- 2a_Ctanθ] (j) ..........( 1 )

so a_A = (a_B)_y (j ) = [- 2a_Ctanθ] (j) .........( 2 )

let N₁ = normal force of B on C , N₂ = normal force between A and B

N₁sinθ = ma_C

mg + N₂ - N₁cosθ = m( a_B )_y = 2ma_Ctanθ

mg - N₂ = m(a_A)_y= 2ma_Ctanθ

so 2mg - N₁cosθ = 4ma_Ctanθ

2mg = ma_Ccotθ + 4ma_Ctanθ

so , a_C = 2gtanθ1+4tan²Î¸

@ Qwerty, plz put on the fbd for that question plz

Kafi pass pahuch gaya hai.. The exact answer is

A_c=\frac{2gsin\theta cos\theta }{1+3sin^{2}\theta} or, \frac{gsin2\theta }{1+3sin^{2}\theta}

arey idiot mahashay(Vivek), answer right hai. itna natak mat karo, exact answer, exam mein answer dekh kar qwerty right answer hi mark karta.

LOL

hahahaha
bechare qwerty ko poora soln. post karna padha!!!
:D

well you have the habit of getting confused when easy questions are asked, so there is no difference if they are asked in matrix match or somewhere else.[3]

If you're solving it, then please let us know. COM concept is involved or not? If yes, Is there any other method for it?

For the first one, I think option (2) and (4) are correct. [1]

for first one t'=0 and t=10

That's right T' must be 0

vivek if u know he answer please post for second one

First is simply, (Mathematical deduction)

T'sin30 = 0 => T' = 0 N
and T+T'cos30=10 => T = 10 N

Well, That was just and refresher and all of you are quite correct. Any Ideas for the second one?

@Euclid, Sorry no friction coeffⁿ is given. The answers are (B,C).

But I don't know how!!

A.)

Find -
1.) The acc_n of Block A w.r.t ground.
2.) The Acc_n of block B w.r.t ground
3.) The Acc_n of Block C w.r.t ground.
And If you want to do more, may be we can add their relative ones.

B.) A mass m is put on an inclined plane of angle theta and tied with a light thread present the upper end of incline (Shown below). Friction coeffⁿ k is present. Now theta is increased from 0 to 90, then

Plot -
1. Tension in thread versus theta
2. Normal on the block vs. theta
3. Friction on the block vs theta
4. Net interaction force between block and Incline versus theta

for 2 i think the coefficient of friction should be given...

P_min = W / √(1+Ï
²)

Ab solve kiya hai toh answer bhi batana tha na. :P . PLZ

i m getting this
a_C = 2gtanθ1+4tan²Î¸

wats d ans given