IIT JEE , AIEEE Forum - Physics , Mechanics

srujana

A block of mass m is placed on a wedge of mass M.The total system is arranged on a smooth horizontal surface.Find the acceleration of the wedge wrt ground and acceleration of the block wrt wedge.

Please derive the relation

relative acceleration of the block = g sin

+ (Mg sin

cos²

)/( M+m sin²

)

tango_goat

accln of wedge w.r.t to ground a=mgsin @ cos@/[M+m(1+sinsq@)]
....resolve forces for m and also consider pseudo force ma for it as it is in non-inertial frame
u can get the req. result

srujana

Iam new to this topic.....plz explain in detail

srujana

plz answer

deepak_agarwal

well this is areally gud problem..note that all u have to be aware of us tht there exists friction in between the masses m and M...make free body diagram of m...then make acceleration diagram of M and m...hen use h concept that a=Ar + A where a is the absolute acc of m....Ar is rel acc of m wrt M and A is the abolute acc of M

swashata4iit

As the block goes down to the wedge the only external force here is the wt of the block or mg. As there is no external force in the X-axis thus the wedge will proceed towards left (from statics) so that the Centre of mass of the system in X-axis remains the same.

Now let

acc. of m w.r.t M be = a = req answer
acc. of M w.r.t ground = a_o .

Now the horizontal component of a is
a_x = acos@ .............(1)

The vertical component of a is
a_y = asin@ ...............(2)

The wedge will provide a normal rxn on the block N and equal force in opposite direction in also exerted on the wedge by the block. (see the FBD in the diagram)

Now frm FBD of the block (m)

The eq of motion along Y-axis

mg - Ncos@ = ma_y ............ (3)

The absolute acc of m along X-axis is (a_x - a_o)

Hence the eq of motion along X-axis

Nsin@ = m(a_x - a_o) ..................(4)

Frmo FBD of the wedge

Nsin@ = Ma_{o ........................(5)}From 4 and 5 we have

a_o = ma_x/(M+m) ........(6)

Again frm 4

Nsin@ = mMa_x/(M+m) .........(7)

Frm 3 we have

Ncos@ = mg - ma_y ...........(8)

by (8)/(7)

cot@ = m(M+m)(g-a_y)/mMa_x

or cot@ = m(M+m)(g/a_x - a_y/a_x)/mM
or cot@ = (g/acos@ - tan@)(M+m)/M

Solve this to get

a = gsin@(M+m)/(M+msin²@) [proved]

As u can see in the question we had to prove that
a = gsin@ + (mgsin@cos²@)/(M+msin²@)

or a = gsin@[M+mcos²@ + msin²@]/(M+msin²@)

hence a = gsin@(M+m)/(M+msin²@)

Plz do rate my efforts

deepak_agarwal

thankx a lott swastha for a wonderful xplaination..u have made the job a lot easier for me..

swashata4iit

Thanks

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