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A block of mass m is placed on a wedge as shown. Find difference between maximum and minimum (a(max) – a(min) ) horizontal rightward acceleration of wedge for which block remains at rest with respect to wedge. (A) 10g/9 (B) g/9 (C) g (D) None of these
Asked by sahuaryanbro1.0 | 01 Jan, 2023, 10:17: PM
Fig.(1) shows the forces acting on block of mass m . Weight mg is resolved as ( mgcos37 ) normal to inclined surface

and ( m g sin37 ) parallel to inclined surface. Where m is mass of block and g is acceleration due to gravity.

When we observe moving wedge and block and if block does not slip, then

the block experience a psuedo force ( m a ) , where a is accceleration of system .

This pseudo force also resolved as ( m a cos37 ) parallel to inclined plane and ( m a sin37) normal to inclined surface.

Normal reaction force N = m ( g cos37 + a sin37 )

friction force μN = (1/3) m ( g cos37 + a sin37 )

where μ =1/3 is friction coefficient

Condition for sliding the block downwards is

( mg sin37 - ma cos37 ) ≥ μN

( mg sin37 - ma cos37 ) ≥ (1/3) m ( g cos37 + a sin37 )

Acceleration a in above expression is minimum acceleration amin of system

we get from above expression , amin = ( g/3 )

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When system acceleration exceeds g/3  , friction force increases because friction depends on normal force N

and normal force N depends on system acceleration a .

if pseudo force component ( ma cos37 ) excceds weight component ( m g sin37) then net force directed upwards

and friction force will be downwards inorder to prevent the block sliding upward.

This situation is shown in fig.(2)

Now, condition for sliding the block upwards is

(ma cos37 - mg sin37 ) ≥ μN

( ma cos37 - mg sin37 ) ≥ (1/3) m ( g cos37 + a sin37 )

Acceleration a in above expression is maximum acceleration amax of system

we get from above expression , amax = ( 13/3 ) g

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Difference ( amax - amin ) = [ (13/3) - (1/3) ] g = 4g

Answered by Thiyagarajan K | 02 Jan, 2023, 10:37: PM

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