A man pulls a crate of weight 20 N along a rough horizontal floor using a string inclined at 60° to the horizontal. The crate is in limiting equilibrium when he pulls with a force of magnitude 4 N. Find the exact normal reaction between the crate and the floor.

(20+2\sqrt3)

(20-4\sqrt3)

(20-2\sqrt3)

(20-2\sqrt2)

Explanation

Resolving vertically, the upward component of the 4 N pull is

4\sin60°=4\times\frac{\sqrt3}{2}=2\sqrt3

N, and this reduces the normal reaction:

R+2\sqrt3=20

, so

R=20-2\sqrt3

N. Adding this component instead of subtracting it would only be correct if the string pulled the crate downward into the floor, giving

20+2\sqrt3

. Using

\sin60°=\sqrt3

, double the correct value, gives

20-4\sqrt3

. Using

\sin45°=\frac{\sqrt2}{2}

in place of

\sin60°

gives

20-2\sqrt2

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

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