Chapter 4.1, Problem 11PP

(a)

To determine

To Draw: A force diagram for the horizontal forces acting on the crate.

Explanation of Solution

Given:

Force along positive x-axis $F_{+x}=317\text{N}i$

Force along negative x-axis $F_{-x}=173\text{N}\left(-i\right)$

Diagram:

  

Figure1. shows the force exerted on the crate due to force applied and due to its weight.

(b)

To determine

To Find: The net force on the crate.

Answer to Problem 11PP

  $F_{net}=\left(144\text{N}\right)i$

Explanation of Solution

Given:

Force along positive x-axis $F_{+x}=317\text{N}i$

Force along negative x-axis $F_{-x}=173\text{N}\left(-i\right)$

Formula used:

Equation of motion:

Net force, ${\displaystyle \sum F}=F_{+x}+F_{-x}$

  $F_{+x}$ = force along positive x-axis

  $F_{-x}$ = force along negative x-axis

Calculation:

Since there is no motion in vertical direction

Therefore

  $N=mg$

Where,

  $N$ = Normal force

  $m$ = mass

  $g$ = gravitational acceleration

Applying equation of motion in horizontal direction

  $\begin{array}{l}{F}_{net}=F_{+x}+F_{-x}\\ F_{net}=\left(317-173\right)i\\ F_{net}=\left(144\text{N}\right)i\end{array}$

Conclusion:

Net force on the crate is $F_{net}=\left(144\text{N}\right)i$ .

(c)

To determine

To Find: The mass of the crate.

Answer to Problem 11PP

  $m=110.77\text{kg}$

Explanation of Solution

Given:

Time, $t=5\text{s}$

Initial velocity, $u=0$

Final velocity, $v=6.5\text{m/s}$

Net force, $F_{net}=\left(144\text{N}\right)$

Formula used:

Equation of motion

  $v=u+at$

Where,

  $v$ is the final velocity

  $u$ is the initial velocity

  $a$ is the acceleration

  $t$ is the time

Calculation:

Using the equation of motion

  $\begin{array}{l}v=u+at\\ 6.5=0+a\times 5\\ a=1.3{\text{m/s}}^2\end{array}$

Force can be given as:

  $\begin{array}{l}F=ma\\ m=\frac{F}{a}\\ m=\frac{144}{1.3}\\ m=110.77\text{kg}\end{array}$

Conclusion:

  $m=110.77\text{kg}$ is the mass of the crate.