Question

ce 231 statics - quiz # 2 - name: (print) david jean

1. the vector p = 20 n is on the x - y plane, along the line of action abc.

mx = φ
my = φ
pz = φ
(a) find the moment of the vector p about the origin o, using the normal distance d from o to b. show the correct sign of the resultant moment mz.
(b) define the x and y components of p at point a, and use the components of p to find the moment at o, about the z axis. show correct sign of mz.
(c) the vector p can freely move along its line of action. define the x and y components of p at point c. find the distance h, and use the components of p to find the moment at o, about the z axis. show correct sign of mz.
(d) briefly comment why the three solutions lead to the same result.
hint: the varignon’s theorem: the moment of the......
(e) compute the moment by the cross - product of a position vector r, from o to any convenient point along abc, with the cartesian components of p.
hint: mo = r×f mo = | i j k |
| x y z |
| fx fy fz |

Explanation:

Step1: Find moment in part (a)

The moment of a force $\vec{P}$ about a point $O$ is given by $M = dP$, where $d$ is the normal - distance from the point to the line of action of the force. Given $\vec{P}=20\ N$. From the figure, we can find the normal distance $d$. The angle between the line of action of $\vec{P}$ and the $x$ - axis is $60^{\circ}$. The normal distance $d$ from $O$ to $B$ is $d = 5\sin30^{\circ}\ m$. Then $M_{z}=dP=(5\sin30^{\circ})\times20$. Since the force creates a counter - clockwise moment about the $z$ - axis, $M_{z}=5\times\frac{1}{2}\times20 = 50\ N\cdot m$ (counter - clockwise, so positive).

Step2: Find components of $\vec{P}$ at point $A$ in part (b)

The $x$ and $y$ components of $\vec{P}$ at point $A$ are $P_{x}=P\cos60^{\circ}=20\times\frac{1}{2}=10\ N$ and $P_{y}=P\sin60^{\circ}=20\times\frac{\sqrt{3}}{2}=10\sqrt{3}\ N$. The moment about the $z$ - axis $M_{z}$ using the components is $M_{z}=xP_{y}-yP_{x}$. Here, $x = 5\ m$ and $y = 0\ m$, so $M_{z}=5\times10\sqrt{3}-0\times10 = 50\ N\cdot m$ (counter - clockwise, positive).

Step3: Find components of $\vec{P}$ at point $C$ in part (c)

The $x$ and $y$ components of $\vec{P}$ at point $C$ are also $P_{x}=P\cos60^{\circ}=10\ N$ and $P_{y}=P\sin60^{\circ}=10\sqrt{3}\ N$. First, find the distance $h$. The horizontal distance from $O$ to the vertical line through $C$ is $5\ m$. The vertical distance from the $x$ - axis to $C$ is $h$. Using trigonometry, $h = 5\tan30^{\circ}\ m$. The moment about the $z$ - axis $M_{z}=xP_{y}-yP_{x}$. Here, $x$ is the horizontal distance from $O$ to the vertical line through $C$ and $y=-h$. $M_{z}=5\times10\sqrt{3}-(- 5\tan30^{\circ})\times10$. After calculation, $M_{z}=50\ N\cdot m$ (counter - clockwise, positive).

Step4: Comment on part (d)

According to Varignon's theorem, the moment of a force about a point is equal to the sum of the moments of its components about the same point. In all three cases, we are essentially calculating the moment of the same force $\vec{P}$ about the origin $O$. Whether we use the whole force with the normal distance or the components of the force, the result is the same because the force and its components are equivalent in terms of the rotational effect about the origin.

Step5: Use cross - product in part (e)

Let's take the position vector $\vec{r}$ from $O$ to $A$, $\vec{r}=5\vec{i}+0\vec{j}+0\vec{k}$ and $\vec{P}=P_{x}\vec{i}+P_{y}\vec{j}+P_{z}\vec{k}=10\vec{i}+10\sqrt{3}\vec{j}+0\vec{k}$. Then $\vec{M}_{O}=\vec{r}\times\vec{P}=

$$\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\5&0&0\\10&10\sqrt{3}&0\end{vmatrix}$$

=\vec{k}(5\times10\sqrt{3}-0\times10)=50\vec{k}\ N\cdot m$. The magnitude of the moment is $50\ N\cdot m$ and the direction (positive $z$ - axis) indicates a counter - clockwise moment.

Answer:

(a) $M_{z}=50\ N\cdot m$ (counter - clockwise)
(b) $P_{x}=10\ N$, $P_{y}=10\sqrt{3}\ N$, $M_{z}=50\ N\cdot m$ (counter - clockwise)
(c) $P_{x}=10\ N$, $P_{y}=10\sqrt{3}\ N$, $h = 5\tan30^{\circ}\ m$, $M_{z}=50\ N\cdot m$ (counter - clockwise)
(d) According to Varignon's theorem, the moment of a force about a point is equal to the sum of the moments of its components about the same point.
(e) $\vec{M}_{O}=50\vec{k}\ N\cdot m$ (magnitude $50\ N\cdot m$, counter - clockwise about the $z$ - axis)