Chapter 8, Problem 98A

(a)

To determine

Force exerted by Hon the surfboard.

Explanation of Solution

Given:

Weight of the surfboard is $143\text{ N}$ .

P lifts one end with a force, $F_P=57\text{ N}$

Formula used:

Since the board has no acceleration,

  $F_{net}=0$

Calculation:

H exerted force must be 86 N

A required diagram is shown above, where $F_H$ is the force exerted by H.

From the above equation,

  $143=F_P+F_H=57+F_H$

So,

  $\begin{array}{l}{F}_H=143-57\\ F_H=86\text{ N}\end{array}$

Conclusion:

H must exert a force of $F_H=86\text{ N}$ .

(b)

To determine

The distance of lifting point of H from the lifting point of P.

Answer to Problem 98A

2.02 m

Explanation of Solution

Given:

Weight of the surfboard is $143\text{ N}$ .

P lifts one end with a force, $F_P=57\text{ N}$

Length of the board is $2.43\text{ m}$ .

Calculation:

The torque due to $F_H$ about P is $F_{H}L$ .

Torque due to weight is $143\left(\frac{2.43}{2}\right)\text{N}\cdot \text{m}$

Here, $\left(\frac{2.43}{2}\right)\text{m}$ is the distance of midpoint of the board from P.

Since the board is not rotating, the sum of moments about P is zero

So,

  $143\left(\frac{2.43}{2}\right)=F_{H}L$

Where, $F_H=86$ N

So,

  $143\left(\frac{2.43}{2}\right)=86L$

  $L=2.02\text{ m}$

Conclusion:

The lifting point by H has a distance of $2.02\text{ m}$ from P.