Wedge Constraint

the simple sol is easy enough

the acc of m is twice of that of M
and the tension on the block is 3t
therefore mgsin@-t=2ma
and 3t=Ma

net work done by internal forces on a system is zero...because positive work and negative work cancel each other...from this we can conclude Tension(along the direction of acc) is inversely proportional to acceleration..

i got acc of the block along the incline is a(3- cos∂ )

yup the acc of block along the incline is a (3 - cos theta)

but how ??

(1)in ground frame, as string doesnt slack , hence work done by string = 0

T(a)+T(a)+T(a)+T(a)(-cosθ)=T.a_m = (T)(a_{m,along incline)}

hence a_{m,along incline} = 3a - acosθ

if u dont get this,

(2) in wedge's frame ,as string doesnt slack in dis frame also , hence work done by string = 0

hence Σ T. a = 0

now in wedge's frame , left wall is having accn a in +ve direction

T(a)+T(a)+T(a) =Ta_m/M

so a_m/M = 3a ..along incline and downwards

or a_m/M = 3acosθ (i) - 3asinθ (j)

a_M = -a(i)

a_m = a_m/M+ a_M = (3acosθ - a) (i) - 3asinθ (j)

now consider a unit vector along the incline , \hat{n}= -cos\theta (i)+sin\theta (j)

now \vec{a_{m}}.\hat{n}= component \; \; of \;\;a_{m}\;\;along\; \; incline

= -(3acosθ-a)(cosθ) - 3asinθ(sinθ) = acosθ - 3a

or 3a-acosθ down the incline

Thanx subhodip and Qwerty [1]