Kinetics: Force, Mass, and Acceleration

FE Fundamentals of Engineering Exam

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Kinetics: Force, Mass, and Acceleration

Key Takeaways Explain This Topic AI

Newton's second law ΣF = ma relates the net external force to mass times acceleration; weight is the special case W = mg.

Solve particle kinetics by pairing a free-body diagram with a kinetic (ma) diagram and applying ΣF = ma along each axis.

For curved paths, ΣFn = mv²/ρ supplies the centripetal force and ΣFt = mat changes the speed.

Rigid-body rotation obeys ΣM_G = I_G α, where I_G is the mass moment of inertia about the center of mass and α is angular acceleration.

Mass moment of inertia depends on mass distribution: a solid disk is mr²/2, a slender rod mL²/12, a solid sphere 2mr²/5.

Rolling without slipping links translation and rotation via vG = rω and aG = rα, and the contact point has zero instantaneous velocity.

Last updated: June 2026

Newton's Second Law for Particles

Kinetics relates the forces on a body to the motion they produce. The central law is: ∑ F ⃗ = m a ⃗ \sum \vec{F} = m\vec{a} ∑ F = m a

resolved into components ΣFx = max and ΣFy = may. Weight is the gravitational special case W = mg with g = 9.81 m/s² (32.2 ft/s²). A reliable solution recipe:

Draw the free-body diagram (all real external forces).

Draw the kinetic diagram (the ma vector in the acceleration direction).

Write ΣF = ma along each axis.

Solve the resulting equations, including any kinematic constraints.

Law Statement
First (inertia) A body keeps its state of rest or uniform motion unless a net force acts
Second (F = ma) Net force equals mass times acceleration
Third (action–reaction) Forces between two bodies are equal and opposite
Worked Example — Curved-Path Force

A 2000 kg car travels at 25 m/s around a flat curve of radius 50 m. The friction force keeping it on the path supplies the centripetal force: ∑ F n = m v 2 ρ = 2000 ( 25 ) 2 50 = 2000 ( 12.5 ) = 25,000 N = 25 kN \sum F_n = m\frac{v^2}{\rho} = 2000\frac{(25)^2}{50} = 2000(12.5) = 25{,}000\text{ N} = 25\text{ kN} ∑ F n = m ρ v 2 = 2000 50 ( 25 ) 2 = 2000 ( 12.5 ) = 25 , 000 N = 25 kN

If the available friction μsN = μs(mg) is less than 25 kN, the car skids — the link between kinematics (v²/ρ) and friction (a Statics idea) is a favorite exam crossover.

Rigid-Body Rotation

When a body rotates, the rotational analog of F = ma applies about the mass center G: ∑ M G = I G α \sum M_G = I_G\,\alpha ∑ M G = I G α

where I_G is the mass moment of inertia about G and α is the angular acceleration (rad/s²). For rotation about a fixed axis O, use ΣM_O = I_O α. The mass moment of inertia measures resistance to angular acceleration and depends on how mass is distributed relative to the axis:

Body (about centroidal axis) I
Slender rod (length L) mL²/12
Solid cylinder / disk (radius r) mr²/2
Thin hollow cylinder (radius r) mr²
Solid sphere (radius r) 2mr²/5
Thin spherical shell (radius r) 2mr²/3
Parallel-Axis Theorem (Mass)

To shift I to a parallel axis a distance d from the centroid: I = I ˉ + m d 2 I = \bar{I} + m d^2 I = I ˉ + m d 2

This is identical in form to the area version but uses mass m , not area. Example: a slender rod of mass m and length L swung about one end has I = mL²/12 + m(L/2)² = mL²/3.

General Plane Motion and Rolling

A rigid body in general plane motion translates and rotates at once, so both equations hold simultaneously: ∑ F ⃗ = m a ⃗ G , ∑ M G = I G α \sum \vec{F} = m\vec{a}_G, \qquad \sum M_G = I_G\,\alpha ∑ F = m a G , ∑ M G = I G α

The body's mass center accelerates per ΣF = ma_G while the body spins per ΣM_G = I_G α.

Rolling Without Slipping

For a wheel or disk that rolls without slipping, the rotation and translation are tied by kinematic constraints: v G = r ω , a G = r α v_G = r\omega, \qquad a_G = r\alpha v G = r ω , a G = r α

The contact point has zero instantaneous velocity — it acts as an instantaneous center of rotation. Static friction at the contact does no work because there is no sliding, which is why energy methods work cleanly for rolling bodies.

Worked Example — Disk Moment of Inertia and Torque

Find the angular acceleration of a solid disk (mass 5 kg, radius 0.3 m) when a 6 N·m torque is applied about its center.

I = mr²/2 = 5(0.3)²/2 = 5(0.09)/2 = 0.225 kg·m² .

Then α = M/I = 6 / 0.225 = 26.7 rad/s² .

Trap — using weight where mass belongs: In SI, ΣF = ma needs mass in kg , not weight in newtons. In US units, mass = W/g in slugs (lb·s²/ft). Mixing weight and mass is the most common kinetics error. A second trap: forgetting that a body rotating about a non-centroidal axis needs the parallel-axis transfer before applying ΣM = Iα.

Constrained Systems and Equations of Motion

Most FE kinetics problems involve connected bodies — blocks linked by a cord over a pulley, or a block on an incline tied to a hanging weight. The method: draw a separate free-body diagram for each body, write ΣF = ma for each (taking the positive direction along the expected motion), and add the kinematic constraint that links their accelerations (for an inextensible cord over an ideal pulley, the two blocks share the same acceleration magnitude). Solving the equations simultaneously gives the common acceleration and the cord tension.

Worked Example — Two-Block System

A 10 kg block on a frictionless table is connected by a cord over a frictionless pulley to a 5 kg hanging block. For the hanging block: 5g − T = 5a. For the table block: T = 10a. Adding: 5g = 15a, so a = 5(9.81)/15 = 3.27 m/s² , and T = 10(3.27) = 32.7 N .

D'Alembert's Principle

An alternative viewpoint treats the inertia term ma as a fictitious "inertia force" acting opposite the acceleration, converting a dynamics problem into a pseudo-statics equilibrium: ΣF − ma = 0. This D'Alembert's principle lets you apply the equilibrium habits from Statics directly. It is optional, but some candidates find the inertial-force bookkeeping clearer for rotating machinery and accelerating-frame problems. Whichever form you use, the physics is identical to ΣF = ma.