Chapter 6.4, Problem 73E

(a)

To determine

To Find:The force required to keep the truck from rolling down the hill

Answer to Problem 73E

Force $30,000\sin {\text{d}}^{\text{o}}\text{lb}$ required to keep the truck from rolling down the hill.

Explanation of Solution

Given information:

A truck with a gross weight of $30,000$ pounds is parked on a slope of $d^{\circ }$

Concept used:

Let $u$ and $v$ be non-zero vectors such that $u=w_1+w_2$

Where $w_1$ and $w_2$ are orthogonal and $w_1$ is parallel to $v$ .

The vector $w_1$ and $w_2$ are vector component of $u$ onto $v$ and is denoted by,

  $\begin{array}{l}{w}_1=pro{j}_v{}^u\\ pro{j}_v{}^u=\left(\frac{u.v}{{\Vert v\Vert }^2}\right)v\end{array}$

Calculation:

The force due to gravity is vertical and downward,

  

Represent the gravitation force by the vector

  $F=-30,000j$

To find the force required to keep the truck from rolling down the slope, project $F$ onto a unit $v$ vector in the direction of the slope, as follows

  $v=\cos d^{\circ }i+\sin d^{\circ }j$

Therefore, the projection of $F$ onto $v$ is,

  $\begin{array}{l}{w}_1=\left(\frac{u.v}{{\Vert v\Vert }^2}\right)v\\ w_1=-30,000\sin d^{\circ }\left[\cos d^{\circ }i+\sin d^{\circ }j\right]\\ \Vert w_1\Vert =\sqrt{{(-300000\sin d^{\circ }\cos d^{\circ })}^2+{(-300000\sin d^{\circ }\sin d^{\circ })}^2}\\ \Vert w_1\Vert =300000\sin d^{\circ }\end{array}$

(b)

To determine

complete the given table

Answer to Problem 73E

d	$0^{\circ }$	$1^{\circ }$	$2^{\circ }$	$3^{\circ }$	$4^{\circ }$	$5^{\circ }$
Force	0	523.6	1047.0	1570.1	2092.7	2614.7

d	$6^{\circ }$	$7^{\circ }$	$8^{\circ }$	$9^{\circ }$	${10}^{\circ }$
Force	3135.9	3656.1	4175.2	4693.0	5209.4

Explanation of Solution

Given information:

Force is given as $30,000\sin \text{ d°lb}$

Calculation:

d	$0°$	$1°$	$2°$	$3°$	$4°$	$5°$
Force	0	523.6	1047.0	1570.1	2092.7	2614.7

d	$6°$	$7°$	$8°$	$9°$	$10°$
Force	3135.9	3656.1	4175.2	4693.0	5209.4

(c)

To determine

To Find:The force perpendicular to the hill when $d=5^{\circ }$

Answer to Problem 73E

Force perpendicular to the hill $\approx 29,885.8\text{ pound}$

Explanation of Solution

Given information:

Force

  $w_1=-30,000\sin d^{\circ }\left[\cos d^{\circ }i+\sin d^{\circ }j\right]$

Concept used:let $u$ and $v$ be nonzero vectors such that $u=w_1+w_2$

where $w_1$ and $w_2$ are orthogonal and $w_1$ is parallel to $v$ .

The vector $w_1$ and $w_2$ are vector component of $u$ onto $v$ and is denoted by

  $\begin{array}{l}{w}_1=pro{j}_v{}^u\\ pro{j}_v{}^u=\left(\frac{u.v}{{\Vert v\Vert }^2}\right)v\end{array}$

Calculation:

Force:

  $w_1=-30,000\sin d^{\circ }\left[\cos d^{\circ }i+\sin d^{\circ }j\right]$

Force perpendicular to the hill $=30,000\cos d^{\circ }$

  $d^{\circ }=5^{\circ }$

Force perpendicular to the hill $=30,000\cos 5^{\circ }$

Force perpendicular to the hill $\approx 29,885.8\text{ pound}$