Question

1.A. Calculate the Normal Force on person A, of mass 84 kg while balancing person B, of mass 50 kg, on his head.B. What would the normal force be if someone handed person B a helium balloon capable of generating a lift force of 300 N?

Answer 1

Given data:

* The mass of person A is,

`m_A=84\text{ kg}`

* The mass of person B is,

`m_B=50\text{ kg}`

Solution:

The diagrammatic representation of the given case is,

The weight of person A is,

`\begin{gathered} w_A=m_Ag_{} \\ w_A=84*9.8 \\ w_A=823.2\text{ N} \end{gathered}`

The weight of person B is,

`\begin{gathered} w_B=m_Bg \\ w_B=50*9.8 \\ w_B=490N_{} \end{gathered}`

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,

`\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}`

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,

`\begin{gathered} w_B=490-300 \\ w_B=190\text{ N} \end{gathered}`

Thus, the magnitude of normal force on person A is,

`\begin{gathered} F_{\text{net}}=w_B \\ F_{\text{net}}=190\text{ N} \end{gathered}`

Hence, the normal force on person A, in this case, is 190 N.