4.8 KiB
4.8 KiB
Aidan Sharpe - Homework 2
1.
a) Point O exists at the origin, and point B exists at (16, 0). A $5$kN force, \vec{F}, makes an angle of \frac{5\pi}{6} with the \hat{x} direction at (12, -15).
a) Find the moment, M_O, about point O.
M_O = \vec{r} \times \vec{F}\vec{r} = (12 - 0)\hat{x} + (-15 - 0)\hat{y} = 12\hat{x} - 15\hat{y}\vec{F} = 5000\cos\left(\frac{5\pi}{6}\right)\hat{x} + 5000\sin\left(\frac{5\pi}{6}\right)\hat{y} = -4330.13\hat{x} + 2500\hat{y}\vec{r} \times \vec{F} =
\begin{vmatrix}
\hat{x} & \hat{y} & \hat{z} \\
12 & -15 & 0 \\
-4330.13 & 2500 & 0 \\
\end{vmatrix} = -34951.91\hat{z}$$
$$M_O = -34951.91\hat{z}\text{[N m]}$$
### b) Find the moment, $M_B$, about point $B$
$$M_B = \vec{r} \times \vec{F}$$
$$\vec{r} = (12 - 16)\hat{x} + (-15 - 0)\hat{y} = -4\hat{x} - 15\hat{y}$$
$$\vec{F} = 5000\cos\left(\frac{5\pi}{6}\right)\hat{x} + 5000\sin\left(\frac{5\pi}{6}\right)\hat{y} = -4330.13\hat{x} + 2500\hat{y}$$
$$\vec{r} \times \vec{F} =
\begin{vmatrix}
\hat{x} & \hat{y} & \hat{z} \\
-4 & -15 & 0 \\
-4330.13 & 2500 & 0 \\
\end{vmatrix} = -74951.91\hat{z}$$
$$M_B = -74951.91\hat{z}\text{[N m]}$$
## 2
A $250$N force is applied at an angle $\frac{5\pi}{12}$ with respect to the $\hat{x}$ direction at a distance $30$mm above and $200$mm to the right of the center of a bolt. Find the moment, $M_B$, about the center of the bolt.
$$M_B = \vec{r} \times \vec{F}$$
Convert distance to meters:
$$\vec{r} = 0.2\hat{x} + 0.03\hat{y}$$
$$\vec{F} = 250\cos\left(\frac{5\pi}{12}\right)\hat{x} + 250\sin\left(\frac{5\pi}{12}\right)\hat{y} = 64.70\hat{x} + 241.48\hat{y}$$
$$\vec{r} \times \vec{F} =
\begin{vmatrix}
\hat{x} & \hat{y} & \hat{z} \\
0.2 & 0.03 & 0 \\
64.70 & 241.48 & 0 \\
\end{vmatrix} = 46.36\hat{z}$$
$$M_B = 46.36\hat{z}\text{[N m]}$$
## 3
Consider a bar made up of two segments, $\overline{AB}$ and $\overline{BC}$, each with length, $1.6$m. Segment $\overline{AB}$ makes an angle $\frac{\pi}{2}$ with the $\hat{x}$ direction, and segment $\overline{BC}$ makes an angle $\frac{3\pi}{4}$ with the $\hat{x}$ direction. A $30$N force, $\vec{P}$, is applied perpendicular to $\overline{BC}$. Determine the moment, $M_B$, about point $B$, and $M_A$ about point $A$.
### a) Find the moment, $M_B$, about point $B$
$$M_B = \vec{r} \times \vec{F}$$
$$\vec{r} = 1.6\cos\left(\frac{3\pi}{4}\right)\hat{x} + 1.6\sin\left(\frac{3\pi}{4}\right)\hat{y} =-1.13\hat{x} + 1.13\hat{y}$$
$$\vec{F} = 30\cos\left(\frac{\pi}{4}\right)\hat{x} + 30\sin\left(\frac{\pi}{4}\right)\hat{y} = 21.21\hat{x} + 21.21\hat{y}$$
$$\vec{r} \times \vec{F} =
\begin{vmatrix}
\hat{x} & \hat{y} & \hat{z} \\
-1.13 & 1.13 & 0 \\
21.21 & 21.21 & 0 \\
\end{vmatrix} = -47.93\hat{z}$$
$$M_B = -47.93\hat{z}\text{[N m]}$$
### b) Find the moment, $M_A$, about point $A$
$$M_A = \vec{r} \times \vec{F}$$
$$\vec{r} = 1.6\cos\left(\frac{3\pi}{4}\right)\hat{x} + \left[1.6 + 1.6\sin\left(\frac{3\pi}{4}\right)\right]\hat{y} = -1.13\hat{x} + 2.73\hat{y}$$
$$\vec{F} = 30\cos\left(\frac{\pi}{4}\right)\hat{x} + 30\sin\left(\frac{\pi}{4}\right)\hat{y} = 21.21\hat{x} + 21.21\hat{y}$$
$$\vec{r} \times \vec{F} =
\begin{vmatrix}
\hat{x} & \hat{y} & \hat{z} \\
-1.13 & 2.73 & 0 \\
21.21 & 21.21 & 0 \\
\end{vmatrix} = -81.87\hat{z}$$
$$M_A = -81.87\hat{z}\text{[N m]}$$
## 4
Consider an arm holding a ball and the moments acting on the elbow. The forearm makes an angle, $-35^\circ$ with the $\hat{x}$ direction. The ball weighs 8lbs, and its center of gravity is a distance of $13$ inches away in the $\hat{x}$ direction. The forearm weighs $5$lbs, and its center of gravity is $6$ inches away along the forearm. A third, balancing force of tension acts $2$ inches down the forearm. Find the force of tension, $\vec{T}$, such that the moment about the elbow, $M_O$, is $0$.
$$M_O = (\vec{r}_T \times \vec{T}) + (\vec{r}_G \times \vec{G}) + (\vec{r}_A \times \vec{A}) = 0$$
Calculate radial vectors:
$$\vec{r}_G = 6\cos(-35^\circ)\hat{x} + 6\sin(-35^\circ)\hat{y} = 4.91\hat{x} - 3.44\hat{y}$$
$$\vec{r}_A = 13\hat{x} + 13\tan(-35^\circ)\hat{y} = 13\hat{x} - 9.10\hat{y}$$
$$\vec{r}_T = 2\hat{x} + 2\tan(-35^\circ)\hat{y} = 2\hat{x} - 1.4\hat{y}$$
Assign force vectors:
$$\vec{G} = -5\hat{y}$$
$$\vec{A} = -8\hat{y}$$
Evaluate known cross products:
$$\vec{r}_G \times \vec{G} =
\begin{vmatrix}
\hat{x} & \hat{y} & \hat{z} \\
4.91 & -3.44 & 0 \\
0 & -5 & 0
\end{vmatrix} = -24.55\hat{z}$$
$$\vec{r}_A \times \vec{A} =
\begin{vmatrix}
\hat{x} & \hat{y} & \hat{z} \\
13 & -9.10 & 0 \\
0 & -8 & 0
\end{vmatrix} = -104\hat{z}$$
Combine all knowns:
$$0 = (\vec{r}_T \times \vec{T}) - 24.55\hat{z} - 104\hat{z}$$
$$\therefore \vec{r}_T \times \vec{T} = 128.55\hat{z}$$
Since $\vec{T}$ only acts in the $\hat{y}$ direction:
$$128.55 = r_{T_x} T_y - r_{T_y} T_x = (2)T_x-(-1.4)(0)$$
$$\therefore 128.55 = (2)T_y$$
$$\therefore T_y = 64.28$$
$$\vec{T} = 64.28\hat{y} \text{[lb in]}$$