Paper 4 · statics - friction

Continuing the crate scenario (weight 20 N, pulling force 4 N at 60° to the horizontal, normal reaction $R=20-2\sqrt3$ N, limiting equilibrium), find the exact coefficient of friction between the crate and the floor.

\dfrac{3+10\sqrt3}{97}

\dfrac{1}{10}

40-4\sqrt3

\dfrac{10+\sqrt3}{97}

Explanation: Resolving horizontally,

F=4\cos60°=2

N. At limiting equilibrium,

F=\mu R

, so

\mu=\dfrac{2}{20-2\sqrt3}=\dfrac{1}{10-\sqrt3}

, and rationalising the denominator gives

\mu=\dfrac{10+\sqrt3}{97}

. Swapping sine and cosine, using

F=4\sin60°=2\sqrt3

for the horizontal component instead of

4\cos60°

, gives

\dfrac{3+10\sqrt3}{97}

after rationalising. Dividing by the crate's weight (20 N) instead of the actual normal reaction gives

\dfrac{2}{20}=\dfrac{1}{10}

. Multiplying

F

and

R

instead of dividing one by the other does not correspond to the friction law at all, giving

40-4\sqrt3

Derived from ZIMSEC Maths Paper 1, June 2008, Q12 (Statistics)

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