Chapter 9, Problem 72GP

(a)

To determine

To Find: The normal force exerted by the floor on each hand.

Answer to Problem 72GP

The normal force exerted by the floor on each hand is $\text{ 258}\text{.8 N}$

Explanation of Solution

Given:

  $\begin{array}{l}\text{Mass of person m = 75 kg}\\ \text{Distance from hand to C}\text{.G}\text{. of body = 40 cm = 0}\text{.4 m}\\ \text{Distance from C}\text{.G}\text{. of body to foot = 95 cm = 0}\text{.95 m}\\ \text{Height of C}\text{.G}\text{. from the floor = 30 cm = 0}\text{.3 m}\end{array}$

Formula used:

First of all, make the equation of forces that are acting in vertical upwards and downwards direction. $\sum {\text{F = F}}_{\text{h}}{\text{ + F}}_{\text{f}}\text{ - Weight of body (W) = 0}$

Then, choosing the Centre of Gravity point as a pivot point and apply the equilibrium conditions to determine the torque on hands and feet.

  $\sum {\text{T = F}}_{\text{h}}\times {\text{ d1 - F}}_{\text{f}}\times \text{d2 = 0}$

Calculation:

Substitute the given values, the equation of forces on hand and foot are as follows,

  $\begin{array}{l}\sum {\text{F = F}}_{\text{h}}{\text{ + F}}_{\text{f}}\text{ - mg = 0}\\ \sum {\text{F = F}}_{\text{h}}{\text{ + F}}_{\text{f}}\text{ - (75}\times \text{9}\text{.8) = 0}\\ \sum {\text{F = F}}_{\text{h}}{\text{ + F}}_{\text{f}}\text{ = 735 N (1)}\end{array}$

Substitute the given values in the equation of torque, the equation of forces on hand and foot are as follows.

  $\begin{array}{l}\sum {\text{T = F}}_{\text{h}}\times \text{ 0}{\text{.4 - F}}_{\text{f}}\times \text{ 0}\text{.95 = 0}\\ {\text{So, F}}_{\text{h}}\times \text{ 0}{\text{.4 = F}}_{\text{f}}\times \text{ 0}\text{.95}\\ {\text{ F}}_{\text{f}}\text{ = }\frac{{\text{F}}_{\text{h}}\times \text{ 0}\text{.4}}{\text{0}\text{.95}}\\ {\text{ F}}_{\text{f}}\text{ = 0}\text{.42}\times {\text{ F}}_{\text{h}}\text{ (2)}\end{array}$

Put the value of equation 2 in equation 1, to obtain the forces on hands.

  $\begin{array}{l}{\text{F}}_{\text{h}}{\text{ + F}}_{\text{f}}\text{ = 735 N}\\ {\text{F}}_{\text{h}}\text{ + 0}{\text{.42 F}}_{\text{h}}\text{ = 735 }\\ \text{1}{\text{.42 F}}_{\text{h}}\text{ = 735}\\ {\text{F}}_{\text{h}}\text{= 517}\text{.6 N (Force on both hands)}\\ {\text{F}}_{\text{f}}\text{ 735 - 517}\text{.6 = 217}\text{.4 (Force on both foot)}\end{array}$

Conclusion:

The normal force exerted by the floor on each hand is $\text{ 258}\text{.8 N}$

b.

To determine

To Find: The normal force exerted by the floor on each foot.

Answer to Problem 72GP

The normal force exerted by the floor on each foot is $\text{108}\text{.7 N}$ .

Explanation of Solution

Given:

  $\begin{array}{l}\text{Mass of person m = 75 kg}\\ \text{Distance from hand to C}\text{.G}\text{. of body = 40 cm = 0}\text{.4 m}\\ \text{Distance from C}\text{.G}\text{. of body to foot = 95 cm = 0}\text{.95 m}\\ \text{Height of C}\text{.G}\text{. from the floor = 30 cm = 0}\text{.3 m}\end{array}$

Formula used:

First of all, make the equation of forces that are acting in vertical upwards and downwards direction. $\sum {\text{F = F}}_{\text{h}}{\text{ + F}}_{\text{f}}\text{ - Weight of body (W) = 0}$

Then, choosing the Centre of Gravity point as a pivot point and apply the equilibrium conditions to determine the torque on hands and feet.

  $\sum {\text{T = F}}_{\text{h}}\times {\text{ d1 - F}}_{\text{f}}\times \text{d2 = 0}$

Calculation:

Substitute the given values, the equation of forces on hand and foot are as follows.

  $\begin{array}{l}\sum {\text{F = F}}_{\text{h}}{\text{ + F}}_{\text{f}}\text{ - mg = 0}\\ \sum {\text{F = F}}_{\text{h}}{\text{ + F}}_{\text{f}}\text{ - (75}\times \text{9}\text{.8) = 0}\\ \sum {\text{F = F}}_{\text{h}}{\text{ + F}}_{\text{f}}\text{ = 735 N (1)}\end{array}$

Substitute the given values in the equation of torque, the equation of forces on hand and foot are as follows.

  $\begin{array}{l}\sum {\text{T = F}}_{\text{h}}\times \text{ 0}{\text{.4 - F}}_{\text{f}}\times \text{ 0}\text{.95 = 0}\\ {\text{So, F}}_{\text{h}}\times \text{ 0}{\text{.4 = F}}_{\text{f}}\times \text{ 0}\text{.95}\\ {\text{ F}}_{\text{f}}\text{ = }\frac{{\text{F}}_{\text{h}}\times \text{0}\text{.4}}{\text{0}\text{.95}}\\ {\text{ F}}_{\text{f}}\text{ = 0}\text{.42}\times {\text{ F}}_{\text{h}}\text{ (2)}\end{array}$

Put the value of equation 2 in equation 1, the forces on hands is obtained.

  $\begin{array}{l}{\text{F}}_{\text{h}}{\text{ + F}}_{\text{f}}\text{ = 735 N}\\ {\text{F}}_{\text{h}}\text{ + 0}{\text{.42 F}}_{\text{h}}\text{ = 735 }\\ \text{1}{\text{.42 F}}_{\text{h}}\text{ = 735}\\ {\text{F}}_{\text{h}}\text{= 517}\text{.6 N (Force on both hands)}\\ {\text{F}}_{\text{f}}\text{ 735 - 517}\text{.6 = 217}\text{.4 (Force on both foot)}\end{array}$

Conclusion:

The normal force exerted by the floor on each foot is $\text{108}\text{.7 N}$ .