Chapter 8.3, Problem 39PP

(a)

To determine

The torque acting on the ladder.

Answer to Problem 39PP

The torque acting on the ladder is ${\tau }_{net}=-0.66\times F_A+0.51\times F_B$ .

Explanation of Solution

Given:

Mass of ladder is 7.3 kg and length 1.92 m.

Formula used:

So, force diagram can be draw as-

Calculation:

Torque acting on ladder

Torque due to $F_A$ and $F_B$

  $\begin{array}{l}{\tau }_A=-0.66\times F_A\left(\text{clock wise direction}\right)\\ {\tau }_B=0.51\times F_B\left(\text{anti clock wise direction}\right)\end{array}$

  $\begin{array}{c}{\tau }_g=0\\ {\tau }_{net}={\tau }_A+{\tau }_B\\ =-0.66\times F_A+0.51\times F_B\end{array}$

Conclusion:

The torque acting on the ladder is ${\tau }_{net}=-0.66\times F_A+0.51\times F_B$ .

(b)

To determine

The equation for rotational equilibrium.

Answer to Problem 39PP

The equation for rotational equilibrium is $0.66\times F_A=0.51\times F_B$ .

Explanation of Solution

Given:

Mass of ladder is 7.3 kg and length 1.92 m.

Formula used:

Equation for rotational equilibrium is

  $F_A+F_B=F_g$

Calculation:

Net Torque acting on ladder

  $\begin{array}{c}{\tau }_{net}={\tau }_A+{\tau }_B=0\\ \Rightarrow -0.66\times F_A+0.51\times F_B=0\\ 0.66\times F_A=0.51\times F_B\end{array}$

Conclusion:

The equation for rotational equilibrium is $0.66\times F_A=0.51\times F_B$ .

(c)

To determine

The equation for rotational equilibrium.

Answer to Problem 39PP

The equation for rotational equilibrium is $F_A=0.436F_g\text{ N}$ .

Explanation of Solution

Given:

Mass of ladder is 7.3 kg and length 1.92 m.

Formula used:

Equation for rotational equilibrium is

  $F_A+F_B=F_g$

Calculation:

Equation for $F_A$ in terms of $F_g$

  $\begin{array}{c}{F}_A+F_B=F_g\mathrm{......}\left(1\right)\\ 0.66\times F_A=0.51\times F_B\\ F_B=\frac{0.66}{0.51}\times F_A\end{array}$

From equation (1)

  $\begin{array}{c}{F}_A+\frac{0.66}{0.51}{F}_A=F_g\\ \left(2.29\right)F_A=F_g\\ \Rightarrow F_A=0.436F_g\text{ N}\end{array}$

Conclusion:

The equation for rotational equilibrium is $F_A=0.436F_g\text{ N}$ .

(d)

To determine

The forces exerted by the two sawhorses change if A were moved very close to, but not directly, the center of mass.

Answer to Problem 39PP

If $r_A$ is decreasing, then $F_A$ will increase from above equation. As sum of $F_A$ and $F_B$ is constant. So, if $F_A$ is increase $F_B$ will decrease.

Explanation of Solution

Given:

Mass of ladder is 7.3 kg and length 1.92 m.

Formula used:

Equation for rotational equilibrium is

  $\begin{array}{l}{F}_A+F_B=F_g\mathrm{......}\left(1\right)\\ r_A{F}_A=r_B{F}_B\mathrm{......}\left(2\right)\end{array}$

  $\begin{array}{l}{F}_A+\frac{r_A}{r_B}{F}_A=F_g\\ F_A=\frac{F_g}{\left(1+\frac{r_A}{r_B}\right)}\\ F_A=\frac{r_B{F}_g}{\left(r_A+r_B\right)}\end{array}$

If $r_A$ is decreasing, then $F_A$ will increase from above equation. As sum of $F_A$ and $F_B$ is constant. So, if $F_A$ is increase $F_B$ will decrease.

Calculation:

Conclusion:

If $r_A$ is decreasing, then $F_A$ will increase from above equation. As sum of $F_A$ and $F_B$ is constant. So, if $F_A$ is increase $F_B$ will decrease.