Engineering Board Exam Reviewer · Engineering Mechanics · easy
A 12 kg block rests on a horizontal surface with a coefficient of friction of 0.25. What is the minimum horizontal force required to start moving the block?
- 470.88
- 3
- 88.29
- 29.43
Answer
D. 29.43
Solution
Weight W = mg = 12 × 9.81 = 117.72 N. Since the surface is horizontal, N = W. Maximum static friction Ff = N = 0.25 × 117.72 = 29.43 N.
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