Chapter 8, Problem 74A

(a)

To determine

The least force exerted by J to lift the board to the horizontal position. Also, the part of the board to be lifted by J to exert the force.

Answer to Problem 74A

The least force applied by J to lift the part of the board on the opposite side of A is $61.25\text{N}$ .

Explanation of Solution

Given info:

The mass $\left(m\right)$ of the board is $12.5\text{kg}$ .

The length of the board is $4.0\text{m}$ .

The one end of the board is held up by A.

Consider the acceleration due to gravity of the Earth is $g=9.80{\text{m/s}}^2$ .

Formula used:

Weight of an is the product of a mass and acceleration due to gravity.

  $W=mg$

Calculation:

Consider a person is holding one end of the board. Then, the least force exerted by another person to lift the other end of the board is equal to half of the weight of the board.

The expression for the least force applied by the other person to lift the board is,

  $F_{\text{least}}=\frac{1}{2}mg$

Substituting the values,

  $\begin{array}{c}{F}_{\text{least}}=\frac{1}{2}\times \left(12.5\text{kg}\right)\left(9.80{\text{m/s}}^2\right)\\ =61.25\text{kg}\cdot {\text{m/s}}^2\times \left(\frac{1\text{N}}{\text{1}\text{kg}\cdot {\text{m/s}}^2}\right)\\ =61.25\text{N}\end{array}$

Conclusion:

Therefore, the least force applied by J to lift the part of the board on the opposite side of A is $61.25\text{N}$ .

(b)

To determine

The greatest force exerted by J to lift the board to the horizontal position. Also, the part of the board to be lifted by J to exert the force.

Answer to Problem 74A

The greatest force applied by J to lift the middle part of the board is $122.5\text{N}$ .

Explanation of Solution

Given:

The mass $\left(m\right)$ of the board is $12.5\text{kg}$ .

The length of the board is $4.0\text{m}$ .

The one end of the board is held up by A.

Consider the acceleration due to gravity of the Earth is $g=9.80{\text{m/s}}^2$ .

Formula used:

Weight of an object is the product of a mass and acceleration due to gravity.

  $W=mg$

Calculation:

In order to apply greatest force in holding a board, the person has to hold the board at the middle of the board.

Show the expression for the greatest force applied by the person to lift the board as follows:

  $F_{\text{greatest}}=mg$

Substituting the values,

  $\begin{array}{c}{F}_{\text{greatest}}=\left(12.5\text{kg}\right)\left(9.80{\text{m/s}}^2\right)\\ =122.5\text{kg}\cdot {\text{m/s}}^2\times \left(\frac{1\text{N}}{\text{1}\text{kg}\cdot {\text{m/s}}^2}\right)\\ =122.5\text{N}\end{array}$

Conclusion:

Therefore, the greatest force applied by J to lift the middle part of the board is $122.5\text{N}$ .