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State Hooke's law and verify it with the help of an experiment. |
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Answer» Solution :Hooke's LAW states that for a small deformation, when the STRESS and strain are proportional to each other. It can be verified in a simple way by stretching a thin straight wire (stretches like spring) of length L and uniform cross sectional area A spring is suspended from a fixed point O. A pan and a pointer are ATTACHED at the free end of the wire as shown in Figure 7.2. Using a vernier scale arrangement the extension produced on the wire is measured. From experiment it is KNOWN that for a given load, the corresponding stretching force is F and the elongation produced on the wire is `DeltaL`.  It is directly proportional to the original length L and inversely proportional to the area of cross section A. A graph is plotted using F on the X-axis and `DeltaL` on the Y-axis. This graph is a straight line passing through the origin as shown in Figure 7.3.  `thereforeDeltaL=(slope)F` Multiplying and dividing by volume, `V=AL`, `F(slope)=(AL)/(AL)DeltaL` REARRANGING, we get `(F)/(A)=((L)/(A(slope)))(DeltaL)/(L)` Therefore, `(F)/(A)prop((DeltaL)/(L))` From the equations, we get Stress, `sigma=("Force")/("Area")=(F)/(A)` Strain, `epsilon=("Change in size")/("Original size")=(Deltal)/(l)` `sigmapropepsilon` i.e., the stress is proportional to the strain in the elastic limit. |
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