M015 · Two contacts. Which one gives?

Stacked Blocks & Friction

Apply a force to either block, test both static-friction constraints and watch the first slip. Isolate either block or the pair to see contact forces cancel and reconcile drive work, motion and frictional loss.

Contact regime $\text{Ready}$

Crate / upper acceleration $\text{Ready}$

Lower acceleration $\text{Ready}$

Friction on upper $\text{Ready}$

Floor / surface friction $\text{Ready}$

Selected-system tangential net force $\text{Ready}$

Replay time $\text{Ready}$

Work ledger residual $\text{Ready}$

ISOLATE / PREDICT / RELEASE

Change the interaction. Follow the consequence.

Signed horizontal drive

F

18.00\,N

Upper mass

m_1

2.00\,kg

Lower mass

m_2

4.00\,kg

Inter-block static coefficient

\mu_{s,1}

0.45\,

Apply force to upper block

b

0.00\,

Floor static coefficient

\mu_{s,0}

0.25\,

Kinetic / static coefficient ratio

r

0.75\,

Gravity

g

9.81\,\frac{\mathrm m}{\mathrm s^2}

All runs start from rest under constant loads. Static friction is whatever equilibrium requires within its bound, not always the maximum. Kinetic coefficients are the corresponding static coefficients times the selected ratio. At equality the solver keeps a feasible sticking contact. Changing controls rebuilds the release from rest.

Upper width 0.8 metres, lower width 2.4 metres. Ideal guides prevent rotation without translational work. Stop before full support is lost or after 2.4 metres of lower travel. No tipping, partial support or post-edge fall.

Force arrows use one fixed gain of 0.03 metres per newton. All arrows start at a body centre as a translated free-body diagram; they do not show contact pressure or torque application points. The selected pair combines external forces at its mass centre. Gold denotes upper/crate, blue lower, pink contact friction, green drive.

The plots and CSV are calculated model references, not acquired detector data or fitted material measurements. The friction model omits speed dependence, heating feedback, compliance, rolling and microscopic stick-slip. The diagram exercise checks force identification; directions and magnitudes are supplied by the model.

Initially resting, guided blocks; stop before overhang. No tipping or post-edge fall.

Start with the essentials

Focus question: Does changing the driven block change the first slip?
One-sentence intuition: Friction can accelerate the upper block in its direction of motion.

Core mathematical model

Balance each body

\sum \vec F_{\mathrm{ext}}=m\vec a

Only interactions on the chosen system belong here.

Static bound, kinetic law

|f_s|\leq\mu_sN,\qquad |f_k|=\mu_kN

Static friction adjusts; sliding friction opposes relative velocity.

Two blocks, one internal pair

m_1a_1=F_1+f,\quad m_2a_2=F_2-f+f_0

Internal friction cancels when both blocks form the system.

Close the work ledger

W_F+W_g=\Delta K+E_{\mathrm{diss}}

Inter-block loss uses relative displacement, not either block displacement alone.

Common difficulties

A bound is not a prescription

Typical misconceptionStatic friction always equals its maximum.

Better mental modelIt takes the value required by a feasible sticking constraint; only sliding invokes the kinetic magnitude.

Run the experiment

01
Choose a system
Switch Upper, Lower and Pair.
What to observe: The external-force list follows the system boundary.
02
Predict first
Compare the all-stick demand against both capacities.
What to observe: A correct diagram need not imply rest.
03
Release and reconcile
Play, pause and scrub the model trajectory.
What to observe: Work, motion and dissipation remain consistent up to the stated travel boundary.