Particle equilibrium
Introducido| English | Significado |
|---|---|
| particle/ˈpɑːtɪkl/ | partícula |
| equilibrium/ˌiːkwɪˈlɪbrɪəm/ | equilibrio |
| concurrent/kənˈkʌrənt/ | concurrente |
The lamp hangs still. What must be true?
- Three forces meet at the lamp ring: the weight and two cable tensions. The ring does not move.
- So the three forces must add to zero. That one fact gives exactly two equations, and two equations find two unknowns.
Two equations for a particle
- A particle 质点 is a body whose size does not matter: every force acts at one point. In equilibrium 平衡, $\Sigma F_x = 0$ and $\Sigma F_y = 0$.
- Draw the free-body diagram first, resolve every force, then write one equation per axis.
A particle can be in equilibrium while it is moving.
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
Zero resultant force means zero acceleration, not zero velocity. A particle moving in a straight line at constant speed is in equilibrium too.
Worked example: the symmetric sign
- A $400\ \text{N}$ sign hangs from two cables, each at $30^{\circ}$ above the horizontal on either side.
- By symmetry both tensions are equal, $T$. Vertically: $2T\sin 30^{\circ} = 400$, so $T = 400\ \text{N}$.
- Horizontally: $T\cos 30^{\circ} - T\cos 30^{\circ} = 0$. The second equation is the check, and it holds.
A $600\ \text{N}$ sign hangs from two symmetric cables, each at $40^{\circ}$ above the horizontal. What is the tension in each cable, in N?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
Vertically $2T\sin 40^{\circ} = 600$, so $T = \dfrac{600}{2 \times 0.643} = 466.7\ \text{N}$. The horizontal parts cancel by symmetry.
Flat cables need huge tension
- Only the vertical parts hold the load: $2T\sin\theta = W$, so $T = \dfrac{W}{2\sin\theta}$. As $\theta$ shrinks, $\sin\theta$ shrinks and $T$ grows.
- For the $400\ \text{N}$ sign: $231\ \text{N}$ at $60^{\circ}$, $400\ \text{N}$ at $30^{\circ}$, $1152\ \text{N}$ at $10^{\circ}$. A perfectly straight cable would need infinite tension, which is why washing lines sag.
The same load hangs from two symmetric cables. At which cable angle above the horizontal is the tension smallest?
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
$T = \dfrac{W}{2\sin\theta}$, and $\sin\theta$ is largest at the steepest angle. At $15^{\circ}$ the tension is almost twice the load.
Worked example: no symmetry
- A $200\ \text{N}$ lamp hangs from cable 1 at $30^{\circ}$ and cable 2 at $45^{\circ}$. Find both tensions.
- $x$: $T_1\cos 30^{\circ} = T_2\cos 45^{\circ}$, so $T_1 = 0.816\,T_2$. $\;y$: $T_1\sin 30^{\circ} + T_2\sin 45^{\circ} = 200$.
- Substitute: $T_2(0.408 + 0.707) = 200$, so $T_2 = 179.3\ \text{N}$ and $T_1 = 146.4\ \text{N}$.
- Check: $146.4\sin 30^{\circ} + 179.3\sin 45^{\circ} = 73.2 + 126.8 = 200\ \text{N}$. The steeper cable carries more.
A $300\ \text{N}$ load hangs from cable 1 at $30^{\circ}$ and cable 2 at $60^{\circ}$ above the horizontal, on opposite sides. What is the tension in cable 2, in N?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
$x$: $T_1\cos 30^{\circ} = T_2\cos 60^{\circ}$, so $T_2 = 1.732\,T_1$. $y$: $0.5\,T_1 + 0.866 \times 1.732\,T_1 = 300$, so $T_1 = 150\ \text{N}$ y $T_2 = 259.8\ \text{N}$. Check: $75 + 225 = 300\ \text{N}$.
Tilt the axes along an unknown
- Choose one axis along an unknown force. That force then drops out of the other equation, and each equation solves one unknown.
- A $500\ \text{N}$ crate on a smooth $20^{\circ}$ slope, held by a rope along the slope. Across the slope: $N = 500\cos 20^{\circ} = 469.8\ \text{N}$. Along it: $T = 500\sin 20^{\circ} = 171.0\ \text{N}$.
Axes along the slope
Along the slope the weight's part is $mg\sin\theta$; across it, $mg\cos\theta$. With friction at zero, a rope along the slope must supply all of $mg\sin\theta$.
An $800\ \text{N}$ trolley is held on a smooth $30^{\circ}$ ramp by a rope parallel to the ramp. What is the rope tension, in N?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
With the $x$-axis along the ramp, the normal force drops out. Along the ramp $T = 800\sin 30^{\circ} = 400\ \text{N}$; across it $N = 800\cos 30^{\circ} = 692.8\ \text{N}$.
Three forces meet at one point
- If three non-parallel forces balance, their lines of action are concurrent 汇交: they pass through one point.
- So two known lines fix the third. That often tells you an unknown direction before any algebra.
Put the method for a particle-equilibrium problem in order.
Dificultad 3/5 · 200 monedas + un pequeño bono en el primer intento de hoy
The diagram comes before the axes, because you choose the axes by looking at where the unknowns point.
Marks that slip away
- Same signs for opposite pulls. Two cables both "go up", but their $x$-components point opposite ways.
- Sharing the load by angle size. Tension follows $\sin\theta$, not the angle itself.
- Assuming symmetry. Use the second equation to check it; never just assume it.
Which statements about particle equilibrium are true? Select all that apply.
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
In the lamp example the steeper $45^{\circ}$ cable carried more ($179.3\ \text{N}$ contra $146.4\ \text{N}$), and tension follows $\sin\theta$ y $\cos\theta$, not the angle.
You've got it
- Particle equilibrium: $\Sigma F_x = 0$ and $\Sigma F_y = 0$; two equations, two unknowns.
- $T = \dfrac{W}{2\sin\theta}$ for a symmetric hang: flatter cables carry far more.
- Put an axis along an unknown to remove it; three balanced forces are concurrent.