Chapter 9, Problem 10TP

To determine

The applied force $F_{\text{i}}$ and an explanation behind the total weight $w$ with the range of values over which the force could exist.

Answer to Problem 10TP

The applied force is around $58\text{N}$ and the forces that should be applied to the wheel barrow should be in range of $14\text{N}$ to $58\text{N}$.

Explanation of Solution

Given:

Number of bricks in the wheel barrow is $20$.

Mass of each brick is $3\text{kg}$.

Formula used:

Write the expression for force.

  $F=mg$

Here, $F$ is the force, $m$ is the mass and $g$ is the acceleration due to gravity.

Write the expression for torque.

  ${\tau }_{\text{net}}=F_{\text{net}}{l}_0$

The mechanical $MA$ advantage of the lever is given as,

  $MA=\frac{l_{\text{i}}}{l_0}$

The torque due to the output force is equal n magnitude and opposite in direction o the torque caused by the weight. Thus,

  $\begin{array}{c}\left(F_{\text{out}}\right)\left(l_0\right)=-\left(F_{\text{net}}\right)\left(l_0\right)\\ F_{\text{out}}=-F_{\text{net}}\end{array}$

Since, the mechanical advantage is the ratio of output force to input force. The ratio can be written as,

  $\frac{F_{\text{out}}}{F_{\text{in}}}=\frac{l_{\text{i}}}{l_0}$

Conclusion:

Force due to wheelbarrow is calculated as,

  $\begin{array}{c}{F}_{\text{wheelbarrow}}=mg\\ =\left(20\text{kg}\right)\left(-9.8\text{m}/{\text{s}}^{\text{2}}\right)\\ =-196\text{N}\end{array}$

Force due to the weight of $20$ bricks is calculated as,

  $\begin{array}{c}{F}_{\text{bricks}}=20\left(mg\right)\\ =20\left(3\text{kg}\right)\left(-9.8\text{m}/{\text{s}}^{\text{2}}\right)\\ =-588\text{N}\end{array}$

Total force due to the weight is calculated as,

  $\begin{array}{c}{F}_{\text{net}}=F_{\text{wheelbarrow}}+F_{\text{bricks}}\\ =\left(-196\text{N}\right)+\left(-588\text{N}\right)\\ =-784\text{N}\end{array}$

The net torque s calculated as,

  $\begin{array}{c}{\tau }_{\text{net}}=F_{\text{net}}{l}_0\\ =\left(-784\text{N}\right)l_0\end{array}$

The output force is calculated as,

  $\begin{array}{c}{F}_{\text{out}}=-F_{\text{net}}\\ =-\left(-784\text{N}\right)\\ =784\text{N}\end{array}$

The input force is calculated as,

  $\begin{array}{c}\frac{F_{\text{out}}}{F_{\text{in}}}=\frac{l_{\text{i}}}{l_0}\\ \frac{784\text{N}}{F_{\text{in}}}=\frac{1.02\text{m}}{0.075\text{m}}\\ F_{\text{in}}=57.65\text{N}\\ F_{\text{in}}\approx 58\text{N}\end{array}$

For an empty wheelbarrow the output force is, $F_{\text{out}}=196\text{N}$

The input force for an empty wheelbarrow is calculated as,

  $\begin{array}{c}\frac{F_{\text{out}}}{F_{\text{in}}}=\frac{l_{\text{i}}}{l_0}\\ \frac{196\text{N}}{F_{\text{in}}}=\frac{1.02\text{m}}{0.075\text{m}}\\ F_{\text{in}}=14.41\text{N}\\ F_{\text{in}}\approx 14\text{N}\end{array}$

Conclusion:

Therefore, the applied force is around $58\text{N}$ and the forces that should be applied to the wheel barrow should be in range of $14\text{N}$ to $58\text{N}$.