Skip to article frontmatterSkip to article content

06 Chain Friction

Aim

Determining the coefficient of static friction.

Subjects

Diagram

.

Figure 1:.

Equipment

Presentation

The chain is laid out straight on a table. One end is slowly pulled over the edge until the chain just does not slip. The coefficient of friction (μs)\left(\mu_{s}\right) between the table top and chain is then μs=l0ll0\mu_{s}=\frac{l_{0}}{l-l_{0}}, where ll is the total length of the chain and l0l_{0} the length of the overhanging portion.

Explanation

No slipping means that forces are in equilibrium: F1=F2F_{1}=F_{2} (see Figure 2).

.

Figure 2:.

The mass of the part of the chain hanging over the edge equals: m1=l0lmm_{1}=\frac{l_{0}}{l} m. This makes: l0F1=l0lmgl_{0} F_{1}=\frac{l_{0}}{l} m g.

The mass of the part of the chain still on the table equals: m2=ll0lmm_{2}=\frac{l-l_{0}}{l} m.

The normal force of that part of the chain equals: FN=ll0lmgF2=μsFN=μsll0lmgF_{N}=\frac{l-l_{0}}{l} m g \rightarrow F_{2}=\mu_{\mathrm{s}} F_{N}=\mu_{s} \frac{l-l_{0}}{l} m g.

F1=F2F_{1}=F_{2} now yields: l0lmg=μsll0lmg\frac{l_{0}}{l} m g=\mu_{s} \frac{l-l_{0}}{l} m g, and so: μs=l0ll0\mu_{s}=\frac{l_{0}}{l-l_{0}}.

Remarks

Sources