a maN stands on a weighing machine placed on horizontal platform. the Machine reads 50 kg by means of suitable mechanism the platform is made to execute harmonic vibrations up and down with a frequency of 2 vibrations per second. what will be the effect on the reading on the weighing machine? The amplitude of vibration of the platform is 5 centimetre. Take g iss equal to 10 M per second square
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Answer:max.reading = 89.5kgf
Min. Reading =10.5kgf
Explanation:m=50kg,v=2Hz,
A=5cm=0.05m=4×9.87×2²×0.05=7.9m/s²
amax=w²A=4 pie ²v²A
Max. Force in the man
=m(g+amax)=50(10+7.9)=895.0N=89.5kgf.
Min. Force in man
=m(g-amax)=50(10-7.9)=105.0N=10.5kgf
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