Force and Acceleration (Newton's Second Law)

NCEES FE Civil spec: Dynamics, C. Force acceleration (e.g., particles, rigid bodies). FE Reference Handbook: Dynamics chapter, kinetics of particles and rigid bodies.

Force-acceleration problems apply Newton's second law: the net force on a body equals its mass times its acceleration. For rigid bodies, the net moment about the center of mass (or a fixed pivot) also equals I times α. Typical FE Civil problems are blocks on floors and inclines, connected bodies over pulleys, braking vehicles and vehicles on curves.

What the NCEES spec covers

Section C, "Force acceleration", covers Newton's second law for particles and rigid bodies, kinetic friction, connected systems, and normal and tangential forces in curvilinear motion.

Key concepts

Free body diagram first. Show weight, normal forces, friction and cable tensions. Then pick axes along and perpendicular to the acceleration.
Kinetic friction during sliding is $μ_{k} N$ and opposes the relative motion.
Connected bodies joined by an inextensible cable share the same acceleration magnitude. Over an ideal pulley, tension is the same on both sides.
Curves. The net force toward the center is $m v^{2} / r$ . For a flat curve, friction supplies it; on a banked curve, the normal force helps.
Rigid bodies need both $\sum F = m a_{G}$ and $\sum M_{G} = I_{G} α$ (or $\sum M_{O} = I_{O} α$ about a fixed pin).

Core formulas

\sum F_{x} = m a_{x} \sum F_{y} = m a_{y} \sum M_{G} = I_{G} α

F_{f} = μ_{k} N \sum F_{n} = \frac{m v ^{2}}{r}

Block sliding down an incline with kinetic friction:

a = g (sin θ - μ_{k} cos θ)

Where it is in the FE Reference Handbook

The Dynamics chapter covers kinetics of particles (Newton's second law in rectangular and normal-tangential coordinates, friction) and plane motion of rigid bodies. The statics friction section gives the friction law.

Worked example

A 10 kg block on a horizontal table is connected by a light cable over a frictionless pulley to a 5 kg block hanging freely. The coefficient of kinetic friction between the table and block is 0.20. Find the acceleration and the cable tension.

Hanging block (down positive):

5 (9.81) - T = 5 a

Table block (toward the pulley positive). The normal force is 10(9.81) = 98.1 N:

T - 0.20 (98.1) = 10 a \Rightarrow T - 19.62 = 10 a

Add the two equations to eliminate T:

49.05 - 19.62 = 15 a \Rightarrow a = 1.96 m/s^{2}

T = 5 (9.81 - 1.962) = 39.2 N

Check with the table block: 19.62 + 10(1.962) = 39.2 N. ✓ The tension is less than the hanging weight (49.05 N), as it must be for the hanging block to accelerate downward.

Common mistakes

Setting the cable tension equal to the hanging weight. That is only true when nothing accelerates.
Using N = mg on an incline or when a pulling force has a vertical component.
Using static friction for a body that is already sliding.
Treating centripetal force as an extra applied force on the FBD. It is the net of the real forces.
Summing moments about an accelerating point other than the center of mass.

See the F = ma example, and the static friction lesson for friction before motion starts.

Practice Problems

Try these from the free PolyCalc FE Civil problem bank. Each one has a full worked solution.

A 20 kg crate on a level floor is pulled by a horizontal 100 N force. The coefficient of kinetic friction is 0.25. The crate's acceleration is most nearly: See the worked solution or practice the Dynamics & Seismic deck
A block slides down a 30° incline with a coefficient of kinetic friction of 0.20. Its acceleration down the slope is most nearly: See the worked solution or practice the Dynamics & Seismic deck
A vehicle travels at 8 m/s around a circular path of radius 20 m. What is its centripetal acceleration? See the worked solution or practice the Dynamics & Seismic deck

Related Calculators and Examples

Frequently Asked Questions

Is cable tension equal to the hanging weight in a pulley problem?

Only if the system is at rest or moving at constant speed. When the hanging mass accelerates downward, the tension is less than its weight.

How do I find acceleration down an incline with friction?

Sum forces along the slope: m g sin(theta) minus the kinetic friction mu_k m g cos(theta) equals m a, so a = g (sin theta - mu_k cos theta).

What equations apply to a rigid body in plane motion?

Sum of forces equals mass times the acceleration of the center of mass in each direction, and the sum of moments about the center of mass equals I times alpha. About a fixed pin you can use the moment of inertia about the pin instead.

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