Solve a Physics Problem: Normal Force Calculation for a 38.0kg and 17.0kg Box

[lmf22]

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?

 

[recon]

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.

 

[Leong]

Upward direction : positive; downward direction : negative
\begin{array}{cc}\\<br /> Consider\ mass\ m_1:\\<br /> Newton&#039;s \ 2nd\ Law\\<br /> \sum \vec{F} =m\vec{a}= 0 \ in\ this\ case\\<br /> Fm_2m_1+(-m_2g)=0\\<br /> Fm_2m_1 = m_2g\\<br /> = 17.0*9.81\\<br /> = 168 \ N\\<br /> Fm_2m_1= - Fm_1m_2 : Newton&#039;s \ 3rd\ Law\\<br /> \end{array}

 

[lmf22]

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