4.2 KiB
Homework 3 - Aidan Sharpe
1
A 50[kg] homogeneous smooth sphere rests on a 30^\circ incline and against a vertical wall.
Since the sphere is in static equilibrium:
\sum \vec{F} = 0
\therefore \sum \vec{F}_x = \sum \vec{F}_y = 0
\sum \vec{F}_y = \vec{F}_g + \vec{F}_{A_y}
\sum \vec{F}_x = \vec{F}_B + \vec{F}_{A_x}
Since \vec{F}_A is normal to the surface at 30^\circ to the -\hat{x} direction:
\vec{F}_{A_x} = F_A \cos(60^\circ)\hat{x}
\vec{F}_{A_y} = F_A \sin(60^\circ)\hat{y}
Find the force due to gravity:
\vec{F}_g = (50)(-9.8)\hat{y} = -490\hat{y}
Solve for \|\vec{F}_A\|:
-490 + F_A \sin(60^\circ) = 0
\therefore \|\vec{F}_A\| = {490 \over \sin(60^\circ)} = 565.8 \text{[N]}
\vec{F}_B + 565.8\cos(60^\circ) = 0
\therefore \vec{F}_B = -282.9\hat{x} \text{[N]}
Since F_{A_x} = -F_B and F_{A_y} = -F_g:
\vec{F}_A = 282.9\hat{x} + 490\hat{y} \text{[N]}
2
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.
Assumptions:
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.
Find F_A and F_B:
\sum \vec{M} = 0 = \vec{M}_g + \vec{M}_A + \vec{M}_B
Find the angle that the pole makes:
\theta = \arccos\left({12 \over 15}\right) = 0.6435
Find the about the cable due to gravity, M_g:
\vec{M}_g = \vec{r}_g \times \vec{F}_g
Since \vec{r}_g is 2.5[m] along the bar from the point of tension:
\vec{r}_g = 2.5 \cos(\theta)\hat{x} + 2.5 \sin(\theta)\hat{y} = 2\hat{x} + 1.5\hat{y}
\vec{F}_g = mg = (150)(-9.8)\hat{y} = -1470\hat{y}
Plug in and evaluate \vec{M}_g:
$$\vec{M}_g =
\begin{vmatrix}
\hat{x} & \hat{y} & \hat{z} \
2 & 1.5 & 0 \
0 & -1470 & 0
\end{vmatrix} = -2940\hat{z}$$
Find the moment about the cable attachment due to the support from point A, \vec{M}_A:
\vec{M}_A = \vec{r}_A \times \vec{F}_A
Since \vec{r}_A is -5[m] along the pole from the point of tension:
\vec{r}_A = -5\cos(\theta)\hat{x} - 5\sin(\theta)\hat{y} = -4\hat{x} - 3\hat{y}
Since \vec{F}_A is unknown, but known to only act in the \hat{x} direction:
$$\vec{M}_A =
\begin{vmatrix}
\hat{x} & \hat{y} & \hat{z} \
-4 & -3 & 0 \
F_A & 0 & 0
\end{vmatrix} = 3\vec{F}_A \hat{z}$$
Find the moment about the cable attachment due to the suport from point B, \vec{M}_B:
\vec{M}_B = \vec{r}_B \times \vec{F}_B
Since \vec{r}_B is 10[m] along the pole from the point of tension:
\vec{r}_B = 10\cos(\theta)\hat{x} + 10\sin(\theta)\hat{y} = 8\hat{x} + 6\hat{y}
Since \vec{F}_B is unknown, but known to only act in the \hat{x} direction:
$$\vec{M}_B =
\begin{vmatrix}
\hat{x} & \hat{y} & \hat{z} \
8 & 6 & 0 \
F_B & 0 & 0
\end{vmatrix} = -6F_B \hat{z}$$
To find the values F_A and F_B:
\sum \vec{M} = 0 = \vec{M}_A + \vec{M}_B + \vec{M}_g
\therefore -2940 + 3F_A - 6F_B = 0
\sum \vec{F}_x = 0 =\vec{F}_A + \vec{F}_B
\therefore\vec{F}_A = -\vec{F}_B
\therefore -2940 = 9\vec{F}_B
\therefore \vec{F}_B = -326.6\hat{x}
\therefore \vec{F}_A = 326.6\hat{x}
4
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.
v^2 = 2a\Delta p = 2(2.75)(3) = 16.5 \text{[m/s]}
v = at
t_\text{decline} = {16.5 \over 2.75} = 6\text{[s]}
\Delta p = vt
t_\text{coasting} = {4 \over 16.5} = 0.242\text{[s]}
t = t_\text{decline} + t_\text{coasting} = 6.242\text{[s]}
5
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.
0^2 = (30)^2 + 2(-9.8)\Delta h
{900 \over (2)(9.8)} = \Delta h = 45.918 \text{[m]}
h = 60 + \Delta h = 105.918 \text{[m]}
\Delta h_\text{final} = 30 + {1 \over 2}(-9.8)t^2
-60 = 30 - 4.9t^2
\therefore t = +\sqrt{{-90 \over -4.9}} = 4.2857 \text{[s]}