The figure shows three blocks in contact and placed on a
smooth horizontal fixed table. The ratio of force exerted by
block A to B and the force by block B to Cis.
- ABC
F1 = 30 N 5 kg 2 kg 3 kg F2 = 10 N
(a) 1:2 (b) 2:3 (c) 3:2 (d) 5:4
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Correct option is
A
3:1
Total mass of three blocks:M=5+2+1=8kg
Acceleration of the blocks: a=
M
F
=
8
16
=2
Let N
AB
and N
BC
be the force exerted by block A on B and by block B on C respectively.
F−N
BA
=m
A
a where N
AB
=N
BA
∴16−N
BA
=5×2=10
⇒N
BA
=6=N
AB
N
AB
−N
CB
=m
B
×a=2×2=4
∴N
CB
=N
BC
=6−4=2
∴
N
BC
N
AB
=
2
6
=
1
3
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