Determine the resultant internal normal and shear force in the member at (a) section a–a and (b) section b–b, each of which passes through point A. The 500-lb load is applied along the centroidal axis of the member.
The formula for the total distance traveled is
\Delta s=\Delta \theta \times r
Therefore, the total distance traveled is
\begin{align*} \Delta s & =\left(200000\:\text{rotations}\:\times \frac{2\pi \:\text{radian}}{1\:\text{rotation}}\right)\left(\frac{1.15\:\text{m}}{2}\right) \\ \Delta s & =722566.3103\:\text{m} \\ \Delta s & =722.6\:\text{km} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
Solution:
The formula for friction is
f=\mu _{k\:}N
When we solve for the normal force, N, in terms of the other variables, we have
N=\frac{f}{\mu _k}
The coefficient of kinetic friction is 0.04. Therefore, the normal force is
\begin{align*} N & =\frac{f}{\mu _k} \\ N & =\frac{0.200\:\text{newton}}{0.04} \\ N & =5.00\:\text{newton} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
Solution:
So, we are given mass, m = 63.0 \ \text{kg} , and acceleration, a = 4.20 \ \text{m/s}^2.
The net force has a formula
\text{F}=\text{m}a
Substituting the given values, we have
\begin{align*} F & = \left( 63.0 \ \text{kg} \right)\left( 4.20 \ \text{m/s}^2 \right) \\ F & = 265 \ \text{kg}\cdot \text{m/s}^2 \\ F & = 265 \ \text{N} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}
Solution:
Part A
The total distance traveled is
\begin{align*} \text{d} & =\left(3\times 120\ \text{m}\right)+\left(1\times 120\:\text{m}\right) \\ \text{d} & =480\:\text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}\\ \end{align*}
Part B
The magnitude of the displacement is
\begin{align*} \text{s }& =\sqrt{\left( s_x \right)^{2\:}+\left( s_y \right)^2} \\ \text{s }& = \sqrt{\left(1\times 120\:\text{m}\right)^2+\left(3\times 120\:\text{m}\right)^2} \\ \text{s }& = 379\ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}
The direction is
\begin{align*} \theta & = \tan^{-1}\left(\frac{s_x}{s_y}\right) \\ \theta & = \tan^{-1}\left(\frac{1\times 120\:\text{m}}{3\times 120 \ \text{m}}\right) \\ \theta & =71.6^{\circ} ,\:\text{E of N} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}
You must be logged in to post a comment.