3-1 (c) Equilibrium under coplanar couples

Example 5:

• A light rod is acted on by a pair of 10 N forces (arm 7 m), a pair of $\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">40</span>}\sqrt{\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">3</span>}}$ N forces (making $\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">6</span>}{\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">0</span>}}^{\circ }$ with a line of 3 m between them), and a pair of 250 N forces (arm 1 m).
• Show that the rod is in equilibrium (given the senses: the first two clockwise, the third anticlockwise).

Solution:

• ${\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">M</span>}}_{\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">1</span>}}=-\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">10</span>}\times \mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">7</span>}=-\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">70</span>}$ N·m.
• ${\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">M</span>}}_{\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">2</span>}}=-\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">40</span>}\sqrt{\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">3</span>}}\times \mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">3</span>}\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">sin</span>}\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">6</span>}{\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">0</span>}}^{\circ }=-\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">40</span>}\sqrt{\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">3</span>}}\times \mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">3</span>}\times {\displaystyle \frac{\sqrt{\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">3</span>}}}{\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">2</span>}}}=-\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">180</span>}$ N·m.
• ${\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">M</span>}}_{\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">3</span>}}=+\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">250</span>}\times \mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">1</span>}=\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">250</span>}$ N·m.
• Sum $=-\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">70</span>}-\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">180</span>}+\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">250</span>}=\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">0</span>}$, so the rod is in equilibrium.

Example 6:

• A light rod $\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">A</span>}\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">B</span>}$, 30 cm long, is pivoted at its midpoint.
• Forces of 7.5 N act at its ends, one up and one down, and a string with tension 10 N pulls at $\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">6</span>}{\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">0</span>}}^{\circ }$ to the rod at a point $\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">C</span>}$.
• What force acting with these keeps the rod horizontal?

Solution:

• The two 7.5 N forces form a couple of moment $-\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">7.5</span>}\times \mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">30</span>}=-\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">225</span>}$ N·cm.
• The additional force must form a couple with the tension. It equals the tension, $\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">F</span>}=\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">10</span>}$ N, in the opposite direction, parallel, and the sum of moments must be $+\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">225</span>}$.
• The arm is $\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">C</span>}\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">D</span>}\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">sin</span>}\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">6</span>}{\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">0</span>}}^{\circ }$ where $\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">D</span>}$ is the point of application along the rod: $\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">10</span>}\times \mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">C</span>}\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">D</span>}\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">sin</span>}\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">6</span>}{\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">0</span>}}^{\circ }=\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">225</span>}$.
• $\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">C</span>}\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">D</span>}={\displaystyle \frac{\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">225</span>}}{\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">10</span>}\times \mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">0.866</span>}}}=\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">15</span>}\sqrt{\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">3</span>}}\approx \mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">26</span>}$ cm, so $\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">D</span>}$ is $\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">15</span>}\sqrt{\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">3</span>}}$ cm from $\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">C</span>}$ along the rod.

Example 7:

• A thin square lamina $\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">A</span>}\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">B</span>}\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">C</span>}\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">D</span>}$ (side 60 cm, weight 200 gf at its centre) is hung from a pin at corner $\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">A</span>}$.
• A couple of moment $\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">3000</span>}\sqrt{\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">2</span>}}$ gf·cm acts in its plane.
• Find the angle $\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">A</span>}\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">C</span>}$ makes with the vertical in equilibrium.

Solution:

• The pin reaction and the weight form a couple (they are equal, opposite and parallel); it must balance the external couple.
• Distance $\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">A</span>}\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">M</span>}$ from the pin to the centre: $\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">A</span>}\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">M</span>}=\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">30</span>}\sqrt{\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">2</span>}}$ cm.
• If $\theta$ is the angle of $\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">A</span>}\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">C</span>}$ with the vertical, the arm of the weight/reaction couple is $\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">A</span>}\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">M</span>}\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">sin</span>}\theta$.
• $\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">200</span>}\times \mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">30</span>}\sqrt{\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">2</span>}}\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">sin</span>}\theta =\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">3000</span>}\sqrt{\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">2</span>}}$, so $\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">sin</span>}\theta ={\displaystyle \frac{\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">3000</span>}\sqrt{\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">2</span>}}}{\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">6000</span>}\sqrt{\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">2</span>}}}}={\displaystyle \frac{\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">1</span>}}{\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">2</span>}}}$.
• Hence $\theta =\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">3</span>}{\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">0</span>}}^{\circ }$ (or $\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">15</span>}{\mathrm{<span\; class="eqx"\; dir="ltr"\; style="direction:ltr;unicode-bidi:isolate;display:inline-block">0</span>}}^{\circ }$ on the other side of the vertical).