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A 100 kg crate rests on a concrete floor where the coefficient of static friction is 0.45. Using g = 9.80 m/s squared, what is the maximum static friction force before the crate starts to slide?
A294 N
B441 N
C980 N
D196 N
Answer & Solution
Correct answer: B. 441 N
1. The normal force on a horizontal floor equals the crate's weight, N = mg = (100 kg)(9.80 m/s squared) = 980 N.
2. The maximum static friction is f_s(max) = mu_s N.
3. Substituting the given coefficient, f_s(max) = (0.45)(980 N) = 441 N.
4. 980 N is just the normal force itself, with no friction coefficient applied, so that option ignores mu_s entirely.
5. 294 N comes from using 0.30 instead of 0.45, which is the kinetic coefficient discussed for the same crate once it is already sliding.
_Source: OpenStax College Physics (CC BY 4.0), Ch 5 "Further Applications of Newton's Laws: Friction, Drag, and Elasticity", section 5.1 Friction_
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