Particle kinetics: the two diagrams
Introducido| English | Significado |
|---|---|
| free-body diagram/friː ˈbɒdi ˈdaɪəɡræm/ | diagrama de cuerpo libre |
| normal force/ˈnɔːml fɔːs/ | the surface's perpendicular push, mg cos beta on an incline rather than mg |
| friction/ˈfrɪkʃn/ | the tangential contact force, at most mu_s N while static and exactly mu_k N once sliding |
| equation of motion | the statement that the force sum equals mass times acceleration |
Forces left, answer right
- Newton's second law is drawn twice: the free-body diagram 受力图 carries every external force, the kinetic diagram carries one vector $m\mathbf{a}$ - never both on the same picture.
- On the certification ramp at 25°, the till's world is weight, normal force 法向力 and friction 摩擦力.
*Forces on one diagram, the answer on the other
Which diagram owns it?
Sort each vector onto the free-body or the kinetic diagram.
On the 25 degree ramp, the normal force equals:
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
The perpendicular equation owns N.
The vector m a belongs on the free-body diagram.
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
It is the answer side - its own kinetic diagram.
The ramp, by the book
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The equation of motion 运动方程 along the slope sets the acceleration; the perpendicular one sets $N$.
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Perpendicular: $N = mg\cos 25° = 22.23$ N - not $mg$.
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Along the slope: $a = g(\sin 25° - 0.20\cos 25°) = 2.368$ m/s², so 1.50 m from rest gives $v = 2.665$ m/s.
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Friction while sliding: $\mu_k N = 4.445$ N.
The 2.5 kg till on a 25 degree ramp: normal force (N, two decimals)?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
N = mg cos 25°.
Same ramp with mu_k = 0.20: acceleration down-slope (m/s², three decimals)?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
a = g(sin - mu cos).
From rest over 1.50 m on that ramp: speed (m/s, three decimals)?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
v = sqrt(2 a s).
The friction force while it slides (N, three decimals)?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
mu_k N = 0.20 × 22.23.
Slide or hold
- On the braking cart the till holds until the cart pulls $\mu_s g = 3.434$ m/s²; the test brakes at 3.0 - margin 12.6%.
- On the turntable friction must supply 7.50 N, so $\mu \geq 0.3058$ against a measured 0.35.
Till on a flat cart, mu_s = 0.35: cart acceleration at which it starts to slide (m/s², three decimals)?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
a = mu_s g.
The turntable demands 7.50 N of friction at 24.53 N normal: minimum mu (three decimals)?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
mu = 7.50 / 24.53.
Three kinetics slips
- Drawing ma as a force. It is the answer, on its own diagram.
- Using mu_s after sliding starts. Static coefficient before, kinetic after.
- Setting N = mg on the incline. The perpendicular equation owns N.
Recap
- Free-body diagram, then axes along and across the motion.
- $N$ from the perpendicular equation, $\mu N$ with the right $\mu$.
- Ramp: $a = 2.368$, $v = 2.665$ m/s; cart slip at 3.434; turntable needs $\mu$ 0.3058.