Laws of MotionMedium
Question
A block of mass m is placed on a wedge. The wedge can be accelerated in four manners marked as (1), (2), (3) and (4) as shown. If the normal reactions in situation (1), (2), (3) and (4) are N1, N2, N3 and N4 respectively and acceleration with which the block slides down on the wedge in situations are b1, b2 b3 and b4 respectively then :




Options
A.N3 > N1 > N2 > N4
B.N4 > N3 > N1 > N2
C.b2 > b3 > b4 > b1
D.b2 > b3 > b1 > b4
Solution
(1) 
Balancing forces perpendicular to incline N = mg cos37° + ma sin37°
N1 =
mg +
ma
and along incline mg sin 37o − ma cos 37o = mb1
b1 =
g −
a
(2)
Similarly for this case get N2 =
mg −
ma
and b2 =
g +
a
N2 =
mg −
ma
(3) Similarly for this case get N3 =
mg +
ma
and b3 =
g +
a
(4) Similarly for this case get N4 =
mg −
ma
and b4 =
g −
a


Balancing forces perpendicular to incline N = mg cos37° + ma sin37°
N1 =
and along incline mg sin 37o − ma cos 37o = mb1
b1 =
(2)

Similarly for this case get N2 =
and b2 =
N2 =
(3) Similarly for this case get N3 =
and b3 =
(4) Similarly for this case get N4 =
and b4 =

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