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Image

In the above image, what would the scale reading be? Let scale reading be $R$.

I was taught that the scale reading will always be the Normal force.

Thus imagining the compression of spring to be $x$ and the spring constant is $K$ I drew the free body diagram and got this:

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Thus I conclude that the reading will show the total mass to be less than ($M_p+m_s$). Is this answer correct? Have I committed any error?

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Consider the person and the spring to constitute a system. The only external forces acting on the system are gravity downward and the normal force $N$ of the scale upward, which keeps the system in equilibrium. So the scale will give you the weight of the spring plus the weight of the person, or $N=(M_{P}+M{s})g$.

Suppose that instead of standing on the spring, you held it. If you then squeezed the spring with your hands, with the same force that you would apply if standing on it. Will the scale reading change?

The force of the person on the spring and spring on the person can be considered internal forces.

Hope this helps.

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  • $\begingroup$ Thankyou very much @Bob D. The kx force was troubling me a lot 😅😅😅 thanks again $\endgroup$ – Mohammad Sanaullah Jan 19 at 17:52
  • $\begingroup$ Then my answer was acceptable? $\endgroup$ – Bob D Jan 19 at 18:46
  • $\begingroup$ Yes yes absolutely Thanks very much 👍🏻👍🏻👍🏻👍🏻 $\endgroup$ – Mohammad Sanaullah Jan 19 at 19:19
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Yes, you have committed an error.

The weighing scale shows the magnitude of the normal force. If you draw a free body diagram of the spring and the person as a single system (by doing that, you do not need to consider the spring compression as they are internal forces), you will see that gravity is the only thing interesting here. So, the whole system is acted upon by gravity, and the only thing balancing against it is the normal force due to the weighing scale.

In that case, you can figure out that $N=(M_p+M_s)g$ is the relation.

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  • $\begingroup$ Is it safe to assume that kx force by the spring acts on both opposite sides of the spring and cancels itself out? $\endgroup$ – Mohammad Sanaullah Jan 19 at 17:54
  • $\begingroup$ @MohammadSanaullah Yes. $\endgroup$ – KV18 Jan 19 at 18:35

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