ReferenceForce, Mass and Acceleration

Formula Sheet

The general laws and relations that every solution in the module is built from. Results the lessons derive for particular cases, such as the acceleration on an incline, the speed limits of a curve or the tension in an Atwood machine, are left out on purpose: derive them from these. Conventions: Hibbeler notation (\(\sum\Fvec = m\avec\), \(\mu_s\), \(\mu_k\), \(a_n = v^2/\rho\)); one positive direction for each axis; cords light and inextensible, pulleys light and smooth; SI units with \(g = 9.81\ \text{m/s}^2\).

Fits on two pages of Letter or A4. For a digital copy, choose “Save as PDF” as the printer.

Newton's second law

Equation of motion

\[ \textstyle\sum\Fvec = m\colMA{\avec} \]

Weight

\[ \colW{W = mg} \]

The resultant of all the forces on a particle; \(\avec\) measured in an inertial frame (the ground). \(1\ \text{N} = 1\ \text{kg}{\cdot}\text{m/s}^2\), \(g = 9.81\ \text{m/s}^2\). Third law: two bodies push or pull on each other with equal, opposite, collinear forces.

Free-body and kinetic diagrams

Procedure: (1) isolate the body; draw every force on it: the weight, and one force at each contact; (2) draw \(m\avec\) on the kinetic diagram, assuming a direction if unknown; (3) choose axes, one along \(\avec\) if you can, and write one equation per direction; (4) kinematics for the motion. A negative answer means the assumed direction was wrong.

\(\colW{W}\) ends at the center of gravity; pushes (\(N\), springs) end on the face they push; friction \(\colF{F_f}\) and cord tensions start where they act.
\(m\avec\) is not a force: never put it on the free-body diagram.

Rectangular components and friction

\[ \textstyle\sum F_x = m\colMA{a_x} \qquad \sum F_y = m\colMA{a_y} \qquad \sum F_z = m\colMA{a_z} \]

No slipping

\[ \colF{F_f} \le \mu_s N \]

Slipping

\[ \colF{F_f} = \mu_k N \]

Does it slip? Assume not; find the \(F_f\) and \(N\) needed; if \(|F_f| \le \mu_s N\) it holds, otherwise redo with \(\mu_k N\) against the slip.

Find \(N\) from the equation across the motion, never by assuming \(N = mg\). On a slope at \(\theta\), the weight has components \(mg\sin\theta\) along it and \(mg\cos\theta\) into it.

Cords and pulleys

One free-body diagram per body. A light cord over light, smooth pulleys has one tension \(T\), the same all along it; it can only pull.

The cord's length, in position coordinates

\[ \textstyle\sum_i c_i\,s_i = \text{const} \quad\Rightarrow\quad \sum_i c_i\,v_i = 0, \quad \sum_i c_i\,a_i = 0 \]

\(s_i\) is measured from a fixed datum along body \(i\)'s line of motion; \(c_i\) is the number of cord segments whose length changes with it.

Kinematics along a line

\[ v = \frac{ds}{dt} \qquad a = \frac{dv}{dt} \qquad a\,ds = v\,dv \]

Constant acceleration only:

\[ v = v_0 + a_c t \] \[ s = s_0 + v_0 t + \tfrac12 a_c t^2 \] \[ v^2 = v_0^2 + 2a_c(s - s_0) \]

Forces that vary

When \(F\) depends on time, position or velocity, so does \(a\): write \(ma = F\), pick the relation that keeps the variable \(F\) depends on and the one you want, separate and integrate.

Separating variables in m a = F
\(F\) depends on	Separate
time	\(m\,dv = F(t)\,dt\)
position	\(m\,v\,dv = F(s)\,ds\)
velocity	\(dt = \dfrac{m\,dv}{F(v)}\) or \(ds = \dfrac{m\,v\,dv}{F(v)}\)

Common force laws: spring \(\colS{F_s = ks}\); fluid drag \(F_D = cv\) (slow) or \(kv^2\) (fast).

Normal and tangential components

\[ \textstyle\sum F_t = m\colMA{a_t} \qquad \sum F_n = m\colMA{a_n} \qquad \sum F_b = 0 \]

\[ a_t = \frac{dv}{dt} = v\frac{dv}{ds} \] \[ a_n = \frac{v^2}{\rho} \] \[ a = \sqrt{a_t^2 + a_n^2} \]

\(t\) along the velocity, \(n\) toward the center of curvature, \(b\) perpendicular to both. \(ma_n\) goes on the kinetic diagram: real forces must add up to it. For a path \(y = f(x)\):

\[ \rho = \frac{\left[1 + (dy/dx)^2\right]^{3/2}}{\left|d^2y/dx^2\right|} \]

Cylindrical components

\[ \textstyle\sum F_r = m(\ddot r - r\dot\theta^2) \] \[ \textstyle\sum F_\theta = m(r\ddot\theta + 2\dot r\dot\theta) \] \[ \textstyle\sum F_z = m\ddot z \]

For a particle guided by a rotating arm or rod, or moving on a path given as \(r(\theta)\). The accelerations come from the Cylindrical Coordinates module.

Systems and choosing a method

\[ \textstyle\sum\Fvec_{\text{ext}} = \sum m_i\avec_i = m\colMA{\avec_G} \] \[ m\rvec_G = \textstyle\sum m_i\rvec_i \]

Internal forces cancel in pairs. Draw the boundary around bodies that move together to find \(a\); cut where you want an internal force, which is then external to each part.

Which method answers which question
You need	Use
a force or an acceleration at an instant	\(\sum\Fvec = m\avec\)
speed at a position	work and energy
velocity after a time; impacts	impulse and momentum

Common mistakes

\(N = mg\) without checking. Angled forces, slopes and curves change it.
\(\mu_k N\) for a body at rest. Static friction is whatever is needed, up to \(\mu_s N\).
\(m\avec\) drawn as a force, or a "centrifugal force" added to the free-body diagram.
\(T = m_B g\) for a hanging block that accelerates.
Constant-acceleration formulas with a force that varies.
Friction in the wrong direction: it opposes the slip, and reverses when the motion does.
km/h not converted to m/s before \(v^2/\rho\).

A solution, step by step

A \(50\ \text{kg}\) crate is pushed up a \(20^\circ\) ramp by a horizontal \(400\ \text{N}\) force, \(\mu_k = 0.2\). Find \(a\).

Diagrams: \(W\), \(N\), \(P\), and friction down the slope (it moves up); \(ma\) up the slope.
Axes: \(x\) up the slope, \(y\) normal to it.
\(\sum F_y = 0\): \(N = 490.5\cos 20^\circ + 400\sin 20^\circ = 597.7\ \text{N}\).
\(\sum F_x = ma\): \(375.9 - 167.8 - 0.2(597.7) = 50a\), so \(a = 1.77\ \text{m/s}^2\).
Check: \(N \gt mg\cos\theta\), as the push presses into the ramp.

The full solution is Example 3.3 in Lesson 3

Check values

Test your method or calculator on these (\(g = 9.81\ \text{m/s}^2\)).

Worked results from the lessons
\(12\ \text{kg}\), \(25^\circ\) slope, \(\mu_k = 0.3\), \(3\ \text{m}\)	\(a = 1.48\ \text{m/s}^2\), \(v = 2.98\ \text{m/s}\)
\(8\ \text{kg}\) on a table (\(\mu_k = 0.2\)), \(4\ \text{kg}\) hanging	\(a = 1.962\ \text{m/s}^2\), \(T = 31.39\ \text{N}\)
\(10\ \text{kg}\), \(P = 20t\), \(\mu_s = 0.5\), \(\mu_k = 0.4\)	slides at \(2.45\ \text{s}\); \(v(4) = 3.91\ \text{m/s}\)
flat curve, \(\rho = 80\ \text{m}\), \(\mu_s = 0.6\)	\(v_{\max} = 21.7\ \text{m/s}\)
banked \(15^\circ\), \(\rho = 120\ \text{m}\), \(\mu_s = 0.3\)	\(v_0 = 17.8\), \(v_{\max} = 27.0\ \text{m/s}\)
\(P = 60\ \text{N}\) on \(2\), \(3\), \(5\ \text{kg}\) in a row	\(F_{AB} = 48\ \text{N}\), \(F_{BC} = 30\ \text{N}\)

Worked in Lessons 3, 4, 5, 6, 7 and 8

Symbols and units

\(\colMA{m\avec}\) mass times acceleration, on the kinetic diagram (N)
\(\colW{W = mg}\) weight (N)
\(N\) normal force (N)
\(\colF{F_f}\) friction force (N)
\(\mu_s,\ \mu_k\) static and kinetic coefficients of friction (no unit)
\(T\) cord tension (N)
\(\rho\) radius of curvature (m)
\(a_n = v^2/\rho\) normal acceleration (m/s²)
\(v_t\) terminal velocity (m/s)

Put the formulas to work in the Practice Lab, test yourself with the Self-Check Quiz, or look up a term in the Glossary.