Question

# All the surfaces are smooth and the strings and pulley's are light. The force exerted by the $$20cm$$ part of the rod on the $$10cm$$ part is

A
6N
B
12N
C
24N
D
36N

Solution

## The correct option is C $$24N$$Mass per unit length = $$\frac{3}{30}kg/cm=0.1kg/cm$$Mass of $$10cm$$ part = $$m_1$$ = $$1kg$$Mass of $$20cm$$ part = $$m_2$$ = $$2kg$$Let $$F$$ be the force of contact betwween themFrom free body diagram$$F-20-a=0$$........(1)$$32-F-2a=0$$..........(2)$$32-20-2a-a=0$$$$12-3a=0$$$$a=\frac{12}{3}=4ms^{-2}$$Contact force $$F=20+a=20+4=24N$$Force exerted = $$24N$$Physics

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