Chapter 9, Problem 71GP

(a)

To determine

To Find: The value of force exerted by each hand.

Answer to Problem 71GP

The value of force on the left hand is $326.67\text{N}$ in an upward direction & the force on the right hand is $228.67\text{N}$ in a downward direction.

Explanation of Solution

Given:

The mass of uniform pole, $m=10\text{kg}$

The length of the pole, $l=2\text{m}$

Formula used:

Let the force on the left hand, $F_L$

Let the force on the right hand, $F_R$

  

Taking torque about point A as zero:

  $\begin{array}{l}{\displaystyle \sum {\tau }_A=0}\\ F_L\times 0.3-mg\times 1=0-(1)\end{array}$

From the equilibrium equation in the vertical direction:

  $F_L+F_R=mg-(2)$

Calculation:

Substitute the given values in equation (1):

  $\begin{array}{l}{F}_L\times 0.3-10(9.8)\times 1=0\\ F_L=\frac{98}{0.3}=326.67\text{N}\end{array}$

Substitute the value of $F_L$ in equation (2):

  $\begin{array}{l}326.67+F_R=10(9.8)\\ F_R=98-326.67=-228.67\text{N}\end{array}$

Conclusion:

The value of force on the left hand is $326.67\text{N}$ in an upward direction & the force on the right hand is $228.67\text{N}$ in a downward direction.

(b)

To determine

To Find: The position of the right hand so that neither hand exert a force greater than 150 N.

Answer to Problem 71GP

0.65m

Explanation of Solution

Given:

The force on each hand, $150\text{N}$

Formula used:

Let the force on the left hand, $F_L$

Let the force on the right hand, $F_R$

  

Taking torque about point A as zero:

  $\begin{array}{l}{\displaystyle \sum {\tau }_A=0}\\ F_L\times x-mg\times 1=0-(1)\end{array}$

Calculation:

Substitute the given values in equation (1):

  $\begin{array}{l}150\times x-10(9.8)\times 1=0\\ x=\frac{98}{150}=0.65\text{m}\end{array}$

Conclusion:

The force will be exerted at point $0.65\text{ m}$ away from point A.

(c)

To determine

To Find: The position of the right hand so that neither hand exert a force greater than 85 N.

Answer to Problem 71GP

0.65m

Explanation of Solution

Given:

The force on each hand, $85\text{N}$

Formula used:

Let the force on the left hand, $F_L$

Let the force on the right hand, $F_R$

  

Taking torque about point A as zero:

  $\begin{array}{l}{\displaystyle \sum {\tau }_A=0}\\ F_L\times x-mg\times 1=0-(1)\end{array}$

Calculation:

Substitute the given values in equation (1):

  $\begin{array}{l}85\times x-10(9.8)\times 1=0\\ x=\frac{98}{85}=1.15\text{m}\end{array}$

Conclusion:

The force will be exerted at point $1.15\text{ m}$ away from point A.