EASY · FRICTION AND DRAG
FRICTION FORCE ON FLAT SURFACE
A wooden crate of mass 8 kg is dragged at constant velocity across a horizontal floor. The coefficient of kinetic friction between crate and floor is μk = 0.4. Find the normal force the floor exerts on the crate, then calculate the kinetic friction force opposing its motion.
§ 01
Step-by-step solution
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Step 1
What is the normal force the floor exerts on the crate?
Hint
On a flat horizontal surface, the normal force exactly balances gravity. There is no vertical acceleration, so F_N = m · g.
Step 2
Given the normal force, what is the kinetic friction force?
Solution walkthrough
Start by drawing a free-body diagram of the crate. Three forces act on it: gravity m·g downward, the normal force F_N upward, and the applied push forward together with kinetic friction backward. Because the crate moves at constant velocity, the vertical forces must cancel — so F_N = m·g = 8 × 9.807 = 78.45 N. This is Amontons' first insight: on a flat surface the normal force is simply the weight.
Once you have F_N, the kinetic friction law — due to Guillaume Amontons, 1699 — gives F_k = μk · F_N directly. With μk = 0.4 this is 0.4 × 78.45 = 31.38 N. Notice that contact area never entered the calculation. Whether the crate sits on its wide face or its narrow face, the friction force is the same — one of the most counterintuitive facts in elementary mechanics, and the result Amontons spent years verifying across every material combination he could test.
The 31.38 N figure also tells you exactly how hard the person pushing must push to keep the crate moving at steady speed: the net horizontal force must be zero, so the applied force must match friction. Any more and the crate accelerates; any less and it slows down.
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