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]}