142 lines
6.2 KiB
TeX
142 lines
6.2 KiB
TeX
\subsection{Rotational Equilibrium}
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This subsection treats the equilibrium case of fixed-axis rotational dynamics. In AP statics, a rigid body remains at rest only when both the translational and rotational balances are satisfied.
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\dfn{Rotational equilibrium and static equilibrium}{Consider a rigid body about a chosen fixed axis with unit vector $\hat{k}$. Let $\vec{\alpha}=\alpha\hat{k}$ denote its angular acceleration, let $\vec{\tau}_{\mathrm{net}}=\tau_{\mathrm{net}}\hat{k}$ denote the net external torque about that axis, and let $\vec{F}_{\mathrm{net}}$ denote the net external force on the body.
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The body is in \emph{rotational equilibrium} if
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\[
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\vec{\alpha}=\vec{0},
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\]
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so its angular velocity is constant. In the common AP statics case, the body is at rest and is also in \emph{static equilibrium}, meaning that both its translational and rotational motion remain unchanged:
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\[
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\vec{F}_{\mathrm{net}}=\vec{0},
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\qquad
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\vec{\alpha}=\vec{0}.
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\]
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Thus rotational equilibrium is the $\alpha=0$ special case of rotational dynamics, and static equilibrium is the at-rest case in which there is also no translational acceleration.}
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\thm{Equilibrium conditions for a rigid body}{Let $\vec{F}_{\mathrm{net}}$ denote the net external force on a rigid body, and let $\vec{\tau}_{\mathrm{net},O}$ denote the net external torque about a chosen point $O$. For static equilibrium in an inertial frame,
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\[
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\sum \vec{F}=\vec{0},
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\qquad
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\sum \vec{\tau}_O=\vec{0}.
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\]
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For a planar fixed-axis problem, choose a positive sense of rotation about the axis. Then the equivalent scalar conditions are
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\[
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\sum F_x=0,
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\qquad
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\sum F_y=0,
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\qquad
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\sum \tau_O=0,
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\]
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where positive and negative torques are assigned by the declared sign convention. Force balance enforces translational equilibrium, and torque balance enforces rotational equilibrium.}
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\nt{In equilibrium problems, the pivot can be chosen wherever it is most convenient. A smart choice is often a point through which one or more unknown forces act, because those forces then contribute zero torque about that point. After using $\sum \tau_O=0$ to solve for a remaining unknown such as a tension, use $\sum \vec{F}=\vec{0}$ to find the support-force components. If $\sum \vec{F}=\vec{0}$, then torque balance about one point is equivalent to torque balance about any other point, so changing the pivot changes the algebra, not the physics. For fixed-axis AP problems, signed scalar torques are fully acceptable once a sign convention such as counterclockwise positive has been declared.}
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\pf{Short explanation from $\tau_{\mathrm{net}}=I\alpha$}{Let $I$ denote the moment of inertia of the rigid body about the chosen fixed axis. Fixed-axis rotational dynamics gives
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\[
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\tau_{\mathrm{net}}=I\alpha.
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\]
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If the body is in rotational equilibrium, then $\alpha=0$, so
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\[
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\tau_{\mathrm{net}}=0.
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\]
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Now let $m$ denote the total mass of the body, and let $\vec{a}_{\mathrm{cm}}$ denote the acceleration of its center of mass. For statics the body also has no translational acceleration, so Newton's second law gives
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\[
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\sum \vec{F}=m\vec{a}_{\mathrm{cm}}=\vec{0}.
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\]
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Therefore a rigid body at rest must satisfy both torque balance and force balance. Conversely, if both balances hold, the body has neither angular acceleration nor translational acceleration, so an initially resting body remains in static equilibrium.}
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\qs{Worked example}{A uniform horizontal beam has length $L=4.0\,\mathrm{m}$ and weight $W_b=200\,\mathrm{N}$. It is hinged to a wall at its left end. A light cable is attached to the right end of the beam and makes an angle $\theta=30^\circ$ above the beam. A sign of weight $W_s=300\,\mathrm{N}$ hangs from the beam at a point $x_s=3.0\,\mathrm{m}$ from the hinge. Let $T$ denote the cable tension. Let the hinge force on the beam be $\vec{H}=H_x\hat{i}+H_y\hat{j}$, where $\hat{i}$ points horizontally to the right and $\hat{j}$ points vertically upward. Choose counterclockwise torque as positive.
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Find:
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\begin{enumerate}[label=(\alph*)]
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\item the cable tension $T$,
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\item the horizontal component $H_x$, and
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\item the vertical component $H_y$ of the hinge force.
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\end{enumerate}}
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\sol Draw the beam's free-body diagram. The external forces on the beam are the hinge force $\vec{H}$ at the left end, the cable force $\vec{T}$ at the right end, the beam's weight $\vec{W}_b$ downward at its center, and the sign's weight $\vec{W}_s$ downward at $x_s=3.0\,\mathrm{m}$.
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Choose the hinge as the pivot. Then the unknown hinge force produces zero torque about that point, which is why this pivot choice is efficient.
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The beam's center is at
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\[
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\frac{L}{2}=2.0\,\mathrm{m}
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\]
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from the hinge. The horizontal component of the cable force acts along the beam, so its line of action passes through the hinge and it produces zero torque about the hinge. Thus only the cable's vertical component contributes to the torque balance:
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\[
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\sum \tau_{\mathrm{hinge}}=0.
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\]
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Using counterclockwise as positive,
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\[
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(T\sin\theta)L-W_b\left(\frac{L}{2}\right)-W_sx_s=0.
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\]
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Substitute the given values:
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\[
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(T\sin 30^\circ)(4.0\,\mathrm{m})-(200\,\mathrm{N})(2.0\,\mathrm{m})-(300\,\mathrm{N})(3.0\,\mathrm{m})=0.
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\]
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Since $\sin 30^\circ=0.50$,
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\[
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(0.50T)(4.0)-400-900=0.
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\]
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So
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\[
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2.0T-1300=0,
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\]
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which gives
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\[
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T=650\,\mathrm{N}.
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\]
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Now apply force balance in the horizontal direction:
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\[
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\sum F_x=0.
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\]
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The cable pulls the beam toward the wall, so its horizontal component is to the left. Therefore,
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\[
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H_x-T\cos 30^\circ=0.
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\]
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Hence
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\[
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H_x=T\cos 30^\circ=(650\,\mathrm{N})(0.866)\approx 5.63\times 10^2\,\mathrm{N}.
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\]
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So the horizontal hinge-force component is
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\[
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H_x\approx 563\,\mathrm{N}
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\]
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to the right.
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Now apply force balance in the vertical direction:
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\[
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\sum F_y=0.
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\]
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Thus
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\[
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H_y+T\sin 30^\circ-W_b-W_s=0.
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\]
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Substitute the values:
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\[
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H_y+(650\,\mathrm{N})(0.50)-200\,\mathrm{N}-300\,\mathrm{N}=0.
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\]
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So
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\[
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H_y+325-500=0,
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\]
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which gives
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\[
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H_y=175\,\mathrm{N}.
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\]
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Thus the vertical hinge-force component is upward.
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The final answers are
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\[
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T=650\,\mathrm{N},
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\qquad
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H_x\approx 563\,\mathrm{N}\text{ to the right},
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\qquad
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H_y=175\,\mathrm{N}\text{ upward}.
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\]
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This is the standard statics strategy: pair $\sum \vec{F}=\vec{0}$ with $\sum \tau=0$, choose a pivot that eliminates unknown torque contributions, solve the torque equation first, and then use force balance to determine the support forces.
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