## physics problem: normal force

A 38.0 kg box (m1) rests on a table. A 17.0 kg box (m2) is placed on top of the 38.0 kg box. What is the normal force that the 38.0 kg box exerts on the 17.0 kg box.

I found that the normal force of just the 38.0kg box is 372.78N and the normal force of just the 17.0kg box is 166.77N. But I don't know how to figure out the normal force that the 38.0kg box exerts on the 17.0kg box.

any help?

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 The 17.0 kg box exerts a force of 166.77 N on the 38.0 kg box. Therefore, the normal force is _______. <--- figure it out yourself.
 Upward direction : positive; downward direction : negative $$\begin{array}{cc}\\ Consider\ mass\ m_1:\\ Newton's \ 2nd\ Law\\ \sum \vec{F} =m\vec{a}= 0 \ in\ this\ case\\ Fm_2m_1+(-m_2g)=0\\ Fm_2m_1 = m_2g\\ = 17.0*9.81\\ = 168 \ N\\ Fm_2m_1= - Fm_1m_2 : Newton's \ 3rd\ Law\\ \end{array}$$ Attached Thumbnails

## physics problem: normal force

Thank you very much guys. Leong, the picture really helps.
Thanks again.