Statics and Dynamics: free practice, theory and problems

1. Statics and equilibrium

What is it about?

Statics deals with bodies at rest (or moving at constant velocity). Then the sum of all forces and the sum of all moments are zero. Almost every structural calculation starts here: before you can find the stresses in a beam, a bolt or a crane arm, you need to know which forces act on it – and many of them, the support reactions, are unknown until you have written the equilibrium equations.

The key tool is the free-body diagram: you "cut" the body free from its surroundings and draw every force the surroundings exert on it.

Concepts and formulas

• A force is a vector. Resolve it: $F_x = F\cos\theta$ and $F_y = F\sin\theta$ ($\theta$ measured from the x-axis).
• Resultant of several forces: $R = \sqrt{(\sum F_x)^2 + (\sum F_y)^2}$.
• Moment about a point O: $M_O = F\cdot d$, where $d$ is the perpendicular distance from O to the line of action of the force (Nm or kNm). Counterclockwise is usually taken as positive.
• Couple: two equal, opposite, parallel forces give a pure moment $M = Fd$, the same about every point.
• A uniformly distributed load $q$ (kN/m) over a length $a$ is replaced by its resultant $qa$ at the middle of the loaded length.
• Supports in 2D: a roller gives 1 unknown (a force perpendicular to the surface), a pin gives 2 ($A_x$, $A_y$), a fixed support gives 3 ($A_x$, $A_y$, $M_A$).
• Statically determinate: as many unknowns as equilibrium equations.

Equilibrium in the plane gives three equations:

How to solve the problems

1. Draw the free-body diagram: the body alone with all external loads, self-weight and support reactions. Assume a direction for each unknown.
2. Resolve inclined forces into x and y components.
3. Take moments about a point where as many unknowns as possible act – they then drop out of the equation.
4. Use $\sum F_x = 0$ and $\sum F_y = 0$ for the rest.
5. Check with an equation you have not used, for example moments about another point. A negative result simply means the force points the other way than assumed.

Example

A 6 m long beam has a pin at A (left end) and a roller at B (right end). It carries 12 kN at 2 m from A and 6 kN at 5 m from A. Find the support reactions.

1. Unknowns: $A_x$, $A_y$ and $B_y$. There are no horizontal loads, so $A_x = 0$.
2. Moments about A, counterclockwise positive:

1. Vertical equilibrium: $A_y + B_y - 12 - 6 = 0$, so $A_y = 18 - 9 = 9$ kN.
2. Check with moments about B: $-A_y\cdot 6 + 12\cdot 4 + 6\cdot 1 = -54 + 54 = 0$. It checks out.

Answer: $A_y = B_y = 9$ kN, both upward, and $A_x = 0$.

Common mistakes

• Forgetting the support reactions or the self-weight in the free-body diagram.
• Using the full arm length instead of the perpendicular distance to the line of action. For a force at an angle $\theta$ to the arm, $M = Fd\sin\theta$.
• Mixing signs: choose the positive sense of rotation once and stick to it.
• Believing the moment about a support is zero because it is a support. Only the contribution of that support's own reaction is zero.
• Mixing kN and m with N and mm in the same equation.

Concepts in this part

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2. Trusses and friction

What is it about?

A truss is a structure made of straight members joined at nodes, for example roof trusses, crane booms, bridges and masts. When the loads act at the joints and the members are pinned at their ends, each member becomes a two-force member: it is either in tension or in compression, and the force acts along the member. That is what makes trusses light and stiff.

Friction is the force that resists sliding between two surfaces. It decides whether a crate stays put on an incline, how much force is needed to drag something, and whether a brake or a bolted joint holds.

Concepts and formulas

• Member force $S$: positive in tension (the member pulls on the joint), negative in compression.
• Method of joints: $\sum F_x = 0$ and $\sum F_y = 0$ at each joint, so at most two unknowns per joint.
• Method of sections: cut through at most three members and use three equilibrium equations for one of the parts.
• Zero-force members: at an unloaded joint with two non-collinear members, both are zero. With three members where two are collinear, the third is zero.
• Static friction: $F \le \mu_s N$. Friction is only as large as equilibrium requires, and sliding starts when $F = \mu_s N$.
• Kinetic friction: $F = \mu_k N$, where usually $\mu_k < \mu_s$.
• Incline at angle $\theta$: $N = mg\cos\theta$, and the component of the weight along the plane is $mg\sin\theta$. The block slides by itself if $\tan\theta > \mu_s$.
• Angle of friction: $\tan\varphi = \mu_s$.
• Centroid of composite areas: $\bar x = \sum A_i x_i / \sum A_i$.

How to solve the problems

1. Find the support reactions for the whole truss, treating it as one rigid body.
2. Start at a joint with at most two unknown member forces. Assume tension in all unknown members.
3. Write $\sum F_x = 0$ and $\sum F_y = 0$ and solve. A negative result means compression.
4. Move on to the next joint with at most two unknowns, or use the method of sections if you only need one particular member.
5. Friction: draw a free-body diagram, find the normal force $N$ from equilibrium perpendicular to the surface (it is not always $mg$), and compare the friction needed with $\mu_s N$.

Example

A triangular truss has a pin at A and a roller at B, 4 m apart. The top joint C is 1.5 m above the midpoint between A and B and carries 12 kN downward. Find the forces in AC and AB.

1. Symmetry gives $A_y = B_y = 6$ kN.
2. Member AC has length $\sqrt{2^2 + 1.5^2} = 2.5$ m, so $\sin\alpha = 1.5/2.5 = 0.6$ and $\cos\alpha = 0.8$.
3. Joint A, vertical: $6 + S_{AC}\cdot 0.6 = 0$, so $S_{AC} = -10$ kN (compression).
4. Joint A, horizontal: $S_{AB} + S_{AC}\cdot 0.8 = 0$, so $S_{AB} = 8$ kN (tension).

Answer: AC and BC carry 10 kN in compression, and AB carries 8 kN in tension. The bottom chord holds the "legs" together so they do not spread.

Common mistakes

• Using $N = mg$ even when the surface is inclined or inclined forces act. Find $N$ from equilibrium.
• Setting friction equal to $\mu_s N$ when the block is at rest and not about to slide. Friction is then only as large as equilibrium requires.
• Mixing up signs of member forces. Assume tension everywhere and let the sign tell you when it is compression.
• Cutting through more than three unknown members.

Concepts in this part

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3. Dynamics

What is it about?

Dynamics describes how things move and why. Kinematics describes the motion itself (position, velocity and acceleration), and kinetics links the motion to the forces through Newton's second law. For an engineer this means braking distances, forces in elevators and cranes, the power a motor must deliver and what happens in a collision.

Three tools cover almost everything: Newton's second law when you want acceleration or forces, the energy method when you want the speed after a certain distance, and the impulse–momentum principle when you want the velocity after an impact or a short force pulse.

Concepts and formulas

• Constant acceleration: $v = v_0 + at$, $s = v_0t + \tfrac12at^2$ and $v^2 = v_0^2 + 2as$.
• Newton's second law: $\sum F = ma$. Draw a free-body diagram first.
• Work $W = Fs\cos\theta$, kinetic energy $E_k = \tfrac12mv^2$ and potential energy $E_p = mgh$.
• Work–energy: $E_{k1} + E_{p1} + W_{other} = E_{k2} + E_{p2}$, where $W_{other}$ is work done by friction, motors and similar. Friction does negative work.
• Power: $P = W/t = Fv$, measured in watts (J/s).
• Momentum $p = mv$. Impulse–momentum: $\int F\,dt = \Delta(mv)$.
• Collisions: momentum is conserved, $m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'$. When the bodies stick together (perfectly inelastic impact), $v' = (m_1v_1 + m_2v_2)/(m_1 + m_2)$.
• Circular motion: $a_n = v^2/r$ toward the center, and $F = mv^2/r$.
• Conversion: 1 m/s = 3.6 km/h.

How to solve the problems

1. Choose the method: forces and acceleration call for Newton, speed after a distance calls for energy, and impacts call for momentum.
2. Draw a free-body diagram and choose a positive direction.
3. Convert to SI units (km/h to m/s, g to kg).
4. Set up the equation and solve for the unknown.
5. Check the order of magnitude and the sign.

Example

A car travels at 90 km/h and brakes hard with locked wheels. The coefficient of friction between tires and road is $\mu_k = 0.7$. How long is the braking distance?

1. Speed: $v_0 = 90/3.6 = 25$ m/s.
2. Newton: friction is the only horizontal force, so $\mu_k mg = ma$ and $a = \mu_k g = 0.7\cdot 9.81 \approx 6.87$ m/s² (deceleration). The mass cancels.
3. Kinematics: $0 = v_0^2 - 2as$ gives

With numbers: $s = 25^2/(2\cdot 0.7\cdot 9.81) \approx 45.5$ m.

1. Check with energy: $\tfrac12mv_0^2 = \mu_k mg\,s$ gives the same expression.

Answer: about 45.5 m. Note that twice the speed gives four times the braking distance.

Common mistakes

• Using km/h in formulas that require m/s.
• Forgetting the weight component or the friction in $\sum F$.
• Believing a net force is needed to keep a constant speed. Constant velocity means $\sum F = 0$.
• Assuming kinetic energy is conserved in an inelastic collision. Only momentum is conserved there.
• Using the constant-acceleration formulas when the acceleration varies.

Concepts in this part

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4. Bending of beams

What is it about?

A beam carries load transverse to its length: floor joists, crane runways, shafts and bridge decks. To size it you need to know how large the internal forces are and where they peak. If you cut the beam at a point $x$, the cut must transmit a shear force $V$ (across the beam) and a bending moment $M$ for each part to stay in equilibrium. The diagrams of $V(x)$ and $M(x)$ show where the beam is loaded hardest – where the moment is largest, the bending stress is largest.

Concepts and formulas

• Sign convention (common): $M$ is positive when the bottom of the beam is in tension, so the beam "smiles".
• Relations: $dV/dx = -q$ and $dM/dx = V$. The moment has a maximum or minimum where $V = 0$.
• A point load gives a jump in $V$ and a kink in $M$. A uniform load gives a linear $V$ and a parabolic $M$.
• Simply supported, point load $P$ at midspan: $M_{max} = PL/4$ and $\delta = PL^3/(48EI)$.
• Simply supported, point load at distance $a$ from A and $b$ from B: $M_{max} = Pab/L$ under the load.
• Simply supported, uniform load $q$: $M_{max} = qL^2/8$ at midspan and $\delta = 5qL^4/(384EI)$.
• Cantilever, point load $P$ at the tip: $M_{max} = PL$ at the fixed end and $\delta = PL^3/(3EI)$.
• Cantilever, uniform load $q$: $M_{max} = qL^2/2$ and $\delta = qL^4/(8EI)$.
• Bending stress: $\sigma = M/W$, where $W = I/y_{max}$. For a rectangle $I = bh^3/12$ and $W = bh^2/6$.
• Superposition: for linear elastic beams the effects of several loads can be added.

How to solve the problems

1. Find the support reactions from equilibrium.
2. Cut the beam at distance $x$ and write equilibrium for the left part. This gives $V(x)$ and $M(x)$.
3. Repeat for each segment between loads and supports, or use the fact that the area under the $V$ diagram equals the change in $M$.
4. Find the largest $|M|$: where $V$ changes sign, under point loads or at the fixed end.
5. If needed, compute the stress $\sigma = M/W$ or the deflection from the table formulas. Use N and mm consistently, with $E$ in N/mm² (MPa).

Example

A simply supported beam of length $L = 6$ m carries a uniform load $q = 5$ kN/m. Find $V(x)$, $M(x)$ and the maximum moment.

1. The total load is $qL = 30$ kN, so $A = B = 15$ kN.
2. Cut at $x$: $V(x) = 15 - 5x$ kN and $M(x) = 15x - 2.5x^2$ kNm.
3. $V = 0$ gives $x = 3$ m, at midspan.
4. $M_{max} = 15\cdot 3 - 2.5\cdot 3^2 = 22.5$ kNm, which matches $qL^2/8 = 5\cdot 36/8 = 22.5$ kNm.
5. At $x = 2$ m, $M = 30 - 10 = 20$ kNm. The moment decreases toward the supports and is zero there.

Answer: $M_{max} = 22.5$ kNm at midspan.

Common mistakes

• Using $qL^2/8$ for a cantilever (it is $qL^2/2$ there), or $PL/4$ when the load is not at midspan.
• Mixing kN, m and mm. In $\delta = PL^3/(48EI)$ everything must be in N and mm (or N and m).
• Forgetting that the moment at a free, unloaded end support is zero.
• Believing the maximum moment is always at midspan. It is where $V = 0$.

Concepts in this part

Practise bending of beams in the app →

5. Work, energy and momentum

What is it about?

Newton's laws can be used directly, but it is often much easier to work with energy and momentum. Then you don't have to follow the motion in detail: you just compare before and after. Questions like "how fast is the cart at the bottom of the hill?" or "how fast do the cars move after the collision?" are solved in a few lines.

Concepts and formulas

• Work: a force $F$ that moves something a distance $s$ does the work $W = F s\cos\theta$, where $\theta$ is the angle between the force and the motion. If the force is perpendicular to the motion, the work is zero.
• Kinetic energy and potential energy:

• Conservation of energy without friction: $E_k + E_p$ is constant. With friction, energy is lost as heat: $W_f = \mu m g\, s$ on a horizontal surface.
• Power is work per time: $P = W/t = F v$ (watts, W).
• Momentum and impulse:

• In an isolated system the total momentum is conserved in all collisions. Kinetic energy is only conserved in perfectly elastic collisions.
• Perfectly inelastic collision (the bodies stick together afterwards):

How to solve the problems

1. Choose a method: if the problem asks about speed and height, use energy. If it is about a collision or a short, strong force, use momentum and impulse.
2. Write down the situation before and after.
3. Energy: $E_{k,before} + E_{p,before} = E_{k,after} + E_{p,after} + W_f$. Momentum: $\sum p_{before} = \sum p_{after}$.
4. Choose a positive direction and keep track of the signs of velocities.

Example

A block slides without friction down a slope that is 2 m high. What is its speed at the bottom?

1. Before: $E_p = mgh$, $E_k = 0$. After: $E_p = 0$, $E_k = \tfrac12 m v^2$.
2. $mgh = \tfrac12 m v^2$ gives $v = \sqrt{2gh} = \sqrt{2\cdot 9.81\cdot 2} \approx 6.26$ m/s.
3. The mass and the angle of the slope do not matter.

Common mistakes

• Using conservation of energy in an inelastic collision. Energy is lost there, use momentum.
• Forgetting the sign: a ball that bounces back changes its momentum by $m(v_1 + v_2)$, not $m(v_1 - v_2)$.
• Computing work for a force that is perpendicular to the motion (the normal force does no work).
• Mixing kJ and J.

Concepts in this part

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