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Q.

A uniform rod of length 60 cm and mass 6kg is acted upon by two forces as shown in the diagram. The force exerted by 45 cm part of the rod on 15 cm part of the rod is

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a

30 N

b

27 N

c

9 N

d

18 N

answer is C.

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Detailed Solution

F = F_{1} + \frac{l_{1}}{L} (F_{2} - F_{1})

$= 45 + \frac{45}{60} (21 - 45)$

$= 45 + \frac{3}{4} (- 24)$

$= 45 - 18$

$= 27 N$

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