Given data:
* The mass of person A is,
[tex]m_A=84\text{ kg}[/tex]* The mass of person B is,
[tex]m_B=50\text{ kg}[/tex]Solution:
The diagrammatic representation of the given case is,
The weight of person A is,
[tex]\begin{gathered} w_A=m_Ag_{} \\ w_A=84\times9.8 \\ w_A=823.2\text{ N} \end{gathered}[/tex]The weight of person B is,
[tex]\begin{gathered} w_B=m_Bg \\ w_B=50\times9.8 \\ w_B=490N_{} \end{gathered}[/tex]The normal force in the upward direction acting on person A due to the ground is equal to the weight of the person.
Thus, the net normal force acting on person A is,
[tex]\begin{gathered} F_{\text{net}}=N-w_A-w_B \\ F_{\text{net}}=823.2-823.2-490 \\ F_{\text{net}}=-490\text{ N} \end{gathered}[/tex]Here, the negative sign indicates the direction of the net force is vertically downward.
Thus, the magnitude of normal force on person A is 490 N.
(B). If person B carries a helium balloon with an upward lift of 300 N, then the weight of person B is decreased to,
[tex]\begin{gathered} w_B=490-300 \\ w_B=190\text{ N} \end{gathered}[/tex]Thus, the magnitude of normal force on person A is,
[tex]\begin{gathered} F_{\text{net}}=w_B \\ F_{\text{net}}=190\text{ N} \end{gathered}[/tex]Hence, the normal force on person A, in this case, is 190 N.
