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If $g=10 N/kg$ , what will show the Scales?

Yes, I know this is a very simple problem. But, I am stuck. $$P=mg \Rightarrow m=\frac{100N+100N}{10N/kg}=20 kg .$$

But, I'm worried about this answer. What's the difference between hand holding the scales and putting them on the table? I think, if the Scales is captured manually, the force applied from above should be meaningless. And answer must be $m=10 kg.$ Is it correct?

Community
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Newuser
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2 Answers2

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The answer is $10 kg$. When you apply Hooke's law to one end of a spring: $F=-kx$, it is implicit that the other end is fixed in place by a force $-F$. This force may be applied by the wall, hand, another mass, etc...

npojo
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The idea is that when you would hang $1\mathrm{kg}$ in a massless basket under a scale, the scale would show that $1\mathrm{kg}$ in the basket. We do know Newtons second law that force equals mass times acceleration: $F=m a$. We know the gravitational acceleration on earth at sea level is approximately $g=10\mathrm{m/s^2}$ this tells us that the force being exerted for $1\mathrm{kg}$ is $10\mathrm{N}$. Now you have to imagine the other way around, when I have a force of $100\mathrm{N}$ pulling on both sides, what would the scale show now? It would show the same conversion but now the other way around so: $m=\frac{F}{a}$ with in this case $a=g$. So $F=200\mathrm{N}$ and $a=10\mathrm{m/s^2}$ gives $m=20\mathrm{kg}$.

Gijsv
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