Vectors and forces
Introducido| English | Significado |
|---|---|
| force/fɔːs/ | fuerza |
| direction/daɪˈrekʃn/ | dirección |
| magnitude/ˈmæɡnɪtjuːd/ | magnitud |
| line of action/laɪn ɒv ˈækʃn/ | línea de acción |
| component/kəmˈpəʊnənt/ | The part of a vector acting in a specified direction. |
| resultant/rɪˈzʌltənt/ | resultante |
| weight/weɪt/ | The gravitational force acting on an object. |
| centre of gravity/ˈsentə ɒv ˈɡrævɪti/ | centro de gravedad |
Two ropes, one crate
- One rope pulls a crate east with $30\ \text{N}$, another pulls north with $40\ \text{N}$. One single pull of $50\ \text{N}$, aimed $53.1^{\circ}$ north of east, would do exactly the same job.
- Swapping several pushes for one, or one push for several, is the first skill of statics. This lesson makes it exact.
A force is a size, a direction and a line
- A force 力 has a magnitude 大小 (in newtons, N), a direction 方向 and a line of action 作用线: the straight line it acts along.
- Measure the angle $\theta$ from the $+x$ axis. Then the components 分量 are $F_x = F\cos\theta$ and $F_y = F\sin\theta$.
- Resolve $100\ \text{N}$ at $30^{\circ}$ above the horizontal.
- $F_x = 100\cos 30^{\circ} = 86.6\ \text{N}$ and $F_y = 100\sin 30^{\circ} = 50.0\ \text{N}$.
- Check by rebuilding: $\sqrt{86.6^{2} + 50.0^{2}} = 100\ \text{N}$. The components are the same force, described another way.
A $250\ \text{N}$ force acts at $40^{\circ}$ above the horizontal. What is its horizontal component, in N?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
$F_x = 250\cos 40^{\circ} = 191.5\ \text{N}$. The vertical part is $250\sin 40^{\circ} = 160.7\ \text{N}$; swapping sine and cosine gives that trap.
If the angle $\theta$ is measured from the $y$-axis, the $x$-component of the force is $F\sin\theta$.
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
The cosine always belongs to the axis the angle is measured from. Measured from $y$, $F_y = F\cos\theta$ y $F_x = F\sin\theta$.
Resultant: add components, then rebuild
- The resultant 合力 is the vector sum. Add all the $F_x$, add all the $F_y$, then rebuild: $R = \sqrt{(\Sigma F_x)^{2} + (\Sigma F_y)^{2}}$.
- Adding numbers on two axes never tangles, even with ten forces. Drawing ten arrows tip-to-tail does.
- A bollard is pulled by three ropes: $200\ \text{N}$ east, $150\ \text{N}$ north and $100\ \text{N}$ west.
- $\Sigma F_x = 200 - 100 = 100\ \text{N}$ and $\Sigma F_y = 150\ \text{N}$.
- $R = \sqrt{100^{2} + 150^{2}} = 180.3\ \text{N}$, at $\tan^{-1}(150/100) = 56.3^{\circ}$ north of east.
Two forces, one resultant
R = F1 + F2
The resultant's components are the sums of the components. Make the two forces cancel and the resultant shrinks to cero.
Three ropes pull a post with $120\ \text{N}$ east, $90\ \text{N}$ north and $60\ \text{N}$ west. What is the magnitude of the resultant, in N?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
$\Sigma F_x = 120 - 60 = 60\ \text{N}$, $\Sigma F_y = 90\ \text{N}$, so $R = \sqrt{60^{2} + 90^{2}} = 108.2\ \text{N}$ (at $56.3^{\circ}$ north of east).
Angles from $+x$ carry their own signs
- Give every force its angle counterclockwise from $+x$. Then $\cos$ and $\sin$ produce the minus signs for you: $100\ \text{N}$ at $210^{\circ}$ is $(-86.6,\ -50.0)\ \text{N}$.
- The calculator's $\tan^{-1}$ only returns angles between $-90^{\circ}$ and $90^{\circ}$. Look at the signs of $\Sigma F_x$ and $\Sigma F_y$ to find the true quadrant.
- Forces of $200\ \text{N}$ at $0^{\circ}$, $150\ \text{N}$ at $90^{\circ}$ and $100\ \text{N}$ at $210^{\circ}$ act at one point.
- $\Sigma F_x = 200 - 86.6 = 113.4\ \text{N}$ and $\Sigma F_y = 150 - 50.0 = 100.0\ \text{N}$.
- $R = 151.2\ \text{N}$ at $41.4^{\circ}$. Both sums are positive, so the first-quadrant angle is correct.
The components of a resultant are $\Sigma F_x = -60\ \text{N}$ y $\Sigma F_y = 80\ \text{N}$. What is its direction, measured counterclockwise from $+x$?
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
The force points left and up, so it lies in the second quadrant. The calculator gives $-53.1^{\circ}$; add $180^{\circ}$ to get $126.9^{\circ}$. The magnitude is $100\ \text{N}$.
Slide along the line, never off it
- On a rigid body a force may slide along its own line of action without changing its effect. Moving it sideways changes how it turns the body.
- A weight 重力 is a force too: $W = mg$ with $g = 9.81\ \dfrac{\text{m}}{\text{s}^{2}}$, acting straight down through the centre of gravity 重心. A $50\ \text{kg}$ block weighs $490.5\ \text{N}$.
What is the weight of a $75\ \text{kg}$ motor, in N? Use $g = 9.81\ \dfrac{\text{m}}{\text{s}^{2}}$.
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
$W = mg = 75 \times 9.81 = 735.8\ \text{N}$, acting straight down through the centre of gravity. Mass is in kg; weight is a force in N.
A force acts on a rigid body. Which changes leave its effect on the body sin cambios? Select all that apply.
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
Sliding along the line and swapping for components describe the same force. A sideways move changes its moment about every point not on the new line.
The tilted-frame check
- Resolve the same forces along axes turned through $45^{\circ}$. The components change, but the rebuilt magnitude must not.
- $300\ \text{N}$ along $+x$ plus $400\ \text{N}$ along $+y$ is $500\ \text{N}$. In the tilted axes it is $(495.0,\ 70.7)\ \text{N}$, and $\sqrt{495.0^{2} + 70.7^{2}} = 500\ \text{N}$ again.
Put the components-first routine for a resultant in order.
Dificultad 3/5 · 200 monedas + un pequeño bono en el primer intento de hoy
Angles first, so the signs come out of $\cos$ y $\sin$ by themselves; the quadrant check comes last because it needs both sums.
Marks that slip away
- Say where the angle is measured from. $F_x = F\cos\theta$ only when $\theta$ is measured from the $x$-axis. Measured from the $y$-axis, $F_x = F\sin\theta$.
- Do not trust the calculator's quadrant. $\Sigma F_x = -60\ \text{N}$, $\Sigma F_y = 80\ \text{N}$ gives $\tan^{-1}(80/{-60}) = -53.1^{\circ}$, but the force points up and left, at $126.9^{\circ}$.
- A resultant needs a direction. "$180\ \text{N}$" alone is half an answer.
Match each term to its meaning.
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
These four words appear in every later unit of the course.
You've got it
- A force has magnitude, direction and line of action; resolve it with $F\cos\theta$ and $F\sin\theta$, $\theta$ from $+x$.
- Resultant: sum the $x$ parts, sum the $y$ parts, rebuild $R$ and find the angle in the right quadrant.
- A force may slide along its line, never across it. Check by rebuilding the magnitude.