108 lines
4.2 KiB
Markdown
108 lines
4.2 KiB
Markdown
# Homework 3 - Aidan Sharpe
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## 1
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A 50[kg] homogeneous smooth sphere rests on a $30^\circ$ incline and against a vertical wall.
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Since the sphere is in static equilibrium:
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$$\sum \vec{F} = 0$$
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$$\therefore \sum \vec{F}_x = \sum \vec{F}_y = 0$$
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$$\sum \vec{F}_y = \vec{F}_g + \vec{F}_{A_y}$$
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$$\sum \vec{F}_x = \vec{F}_B + \vec{F}_{A_x}$$
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Since $\vec{F}_A$ is normal to the surface at $30^\circ$ to the $-\hat{x}$ direction:
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$$\vec{F}_{A_x} = F_A \cos(60^\circ)\hat{x}$$
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$$\vec{F}_{A_y} = F_A \sin(60^\circ)\hat{y}$$
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Find the force due to gravity:
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$$\vec{F}_g = (50)(-9.8)\hat{y} = -490\hat{y}$$
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Solve for $\|\vec{F}_A\|$:
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$$-490 + F_A \sin(60^\circ) = 0$$
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$$\therefore \|\vec{F}_A\| = {490 \over \sin(60^\circ)} = 565.8 \text{[N]}$$
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$$\vec{F}_B + 565.8\cos(60^\circ) = 0$$
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$$\therefore \vec{F}_B = -282.9\hat{x} \text{[N]}$$
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Since $F_{A_x} = -F_B$ and $F_{A_y} = -F_g$:
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$$\vec{F}_A = 282.9\hat{x} + 490\hat{y} \text{[N]}$$
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## 2
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A uniform 150[kg], 15[m] long pole is supported by vertical walls spaced 12[m] apart at points $A$ and $B$. A vertical tension force is applied 5[m] from point $A$ (10[m] from point $B$). Find the reactions at $A$ and $B$.
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#### Assumptions:
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The moment, $M_g$, acts at the center of mass. The pole pivots around the point where the cable is attached. The forces at points $A$ and $B$ act strictly in the horizontal direction. There is no friction between the pole and the walls.
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#### Find $F_A$ and $F_B$:
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$$\sum \vec{M} = 0 = \vec{M}_g + \vec{M}_A + \vec{M}_B$$
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Find the angle that the pole makes:
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$$\theta = \arccos\left({12 \over 15}\right) = 0.6435$$
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Find the about the cable due to gravity, $M_g$:
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$$\vec{M}_g = \vec{r}_g \times \vec{F}_g$$
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Since $\vec{r}_g$ is 2.5[m] along the bar from the point of tension:
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$$\vec{r}_g = 2.5 \cos(\theta)\hat{x} + 2.5 \sin(\theta)\hat{y} = 2\hat{x} + 1.5\hat{y}$$
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$$\vec{F}_g = mg = (150)(-9.8)\hat{y} = -1470\hat{y}$$
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Plug in and evaluate $\vec{M}_g$:
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$$\vec{M}_g =
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\begin{vmatrix}
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\hat{x} & \hat{y} & \hat{z} \\
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2 & 1.5 & 0 \\
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0 & -1470 & 0
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\end{vmatrix} = -2940\hat{z}$$
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Find the moment about the cable attachment due to the support from point $A$, $\vec{M}_A$:
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$$\vec{M}_A = \vec{r}_A \times \vec{F}_A$$
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Since $\vec{r}_A$ is -5[m] along the pole from the point of tension:
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$$\vec{r}_A = -5\cos(\theta)\hat{x} - 5\sin(\theta)\hat{y} = -4\hat{x} - 3\hat{y}$$
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Since $\vec{F}_A$ is unknown, but known to only act in the $\hat{x}$ direction:
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$$\vec{M}_A =
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\begin{vmatrix}
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\hat{x} & \hat{y} & \hat{z} \\
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-4 & -3 & 0 \\
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F_A & 0 & 0
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\end{vmatrix} = 3\vec{F}_A \hat{z}$$
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Find the moment about the cable attachment due to the suport from point $B$, $\vec{M}_B$:
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$$\vec{M}_B = \vec{r}_B \times \vec{F}_B$$
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Since $\vec{r}_B$ is 10[m] along the pole from the point of tension:
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$$\vec{r}_B = 10\cos(\theta)\hat{x} + 10\sin(\theta)\hat{y} = 8\hat{x} + 6\hat{y}$$
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Since $\vec{F}_B$ is unknown, but known to only act in the $\hat{x}$ direction:
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$$\vec{M}_B =
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\begin{vmatrix}
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\hat{x} & \hat{y} & \hat{z} \\
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8 & 6 & 0 \\
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F_B & 0 & 0
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\end{vmatrix} = -6F_B \hat{z}$$
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To find the values $F_A$ and $F_B$:
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$$\sum \vec{M} = 0 = \vec{M}_A + \vec{M}_B + \vec{M}_g$$
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$$\therefore -2940 + 3F_A - 6F_B = 0$$
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$$\sum \vec{F}_x = 0 =\vec{F}_A + \vec{F}_B$$
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$$\therefore\vec{F}_A = -\vec{F}_B$$
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$$\therefore -2940 = 9\vec{F}_B$$
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$$\therefore \vec{F}_B = -326.6\hat{x}$$
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$$\therefore \vec{F}_A = 326.6\hat{x}$$
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## 4
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If the car accelerates at 2.75[$\text{m}/\text{s}^2$] for 3[m] and then maintains speed for 4[m], find the time it takes to travel the entire distance.
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$$v^2 = 2a\Delta p = 2(2.75)(3) = 16.5 \text{[m/s]}$$
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$$v = at$$
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$$t_\text{decline} = {16.5 \over 2.75} = 6\text{[s]}$$
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$$\Delta p = vt$$
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$$t_\text{coasting} = {4 \over 16.5} = 0.242\text{[s]}$$
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$$t = t_\text{decline} + t_\text{coasting} = 6.242\text{[s]}$$
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## 5
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A ball is thrown upwards with an initial velocity of 30[m/s] at the edge of a 60[m] high cliff. Find the maximum height above the ground, $h$, and the total time, $t$, before the ball hits the ground.
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$$0^2 = (30)^2 + 2(-9.8)\Delta h$$
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$${900 \over (2)(9.8)} = \Delta h = 45.918 \text{[m]}$$
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$$h = 60 + \Delta h = 105.918 \text{[m]}$$
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$$\Delta h_\text{final} = 30 + {1 \over 2}(-9.8)t^2$$
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$$-60 = 30 - 4.9t^2$$
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$$\therefore t = +\sqrt{{-90 \over -4.9}} = 4.2857 \text{[s]}$$
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