Question

Wedge C is fixed to the ground. Block B is placed on wedge A which slides down wedge C.
Mass of A = 15kg
Mass of B = 6kg

Find acceleration of ground of block B with respect to i) ground ii) wedge A

Answer 1

ANSWER

T(1+cosθ)−Nsinθ)=Ma .....(i)

FBD of wedge from ground frame

mgsinθ−macosθ−T=ma

⇒mgsinθ−T=ma(1+cosθ) .......(ii)

N=m(gcosθ+asinθ) ......(iii)

Using (i) + (ii) (1+cosθ)+(iii)sinθ

mgsinθ+mgsinθcosθ=

Ma+ma(1+cosθ)

2

+mgsinθcosθ+masin

2

θ

⇒a=

M+2m(1+cosθ)

mgsinθ

given θ=37

o

,m=5kg and M=32kg

so, a=

5

3

m/s

2

Answer 2

Answer:

Explanation:

this can be solved by multiple methods

first i will say system method

lets assume that block a moves down the incline and block b moves right wards with respect to block a

summation fx=(M+m)gsin30

now here we want acceleration along and perpendicular to incline

so take components with respect to that

so it becomes Acostheta along incline and Asintheta perpendicular down the incline

lets take our axis as right and up along the incline surface as our axis

then acceleration of block b with respect to block a = Acostheta-Asintheta

acceleration of block b with respect to ground if acceleration of block a is B with respect to ground

therefore acceleration of block b with respect to ground = Acostheta-B

and acceleration of block a = -B

therefore

M(-B)+m(Acostheta-B)  =( M+m)gsintheta

now apply pseudo along acceleration of block b with respect to block a

you will be able to get the relation

Bcostheta=A

M(-B)+m(Bcos^2 theta -B) =( M+m)gsintheta

M(-B)+m(-Bsin^2 theta ) = (M+m)gsintheta

taking magnitude

B=(m+M)gsintheta / M+msin^2 theta

you can see that block b is not moving with respect to ground in vertical axis

Taking ground as the axis

Acceleration of block b will be  = (-Bcostheta+A)i-Bsintheta j

there will be motion only in vertical direction

so acceleration of block b will be  = Bsintheta

acceleration of block b with respect to ground =( M+m)gsin^2 theta / M+msin^2 theta