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A 15.18 Kg box is at rest on a table. What is the applied force needed to just overcome the force of static friction if the coefficient of friction is 0.309? Round your answer to 2 decimal places.A 15.18 Kg box is at rest on a table. What is the applied force needed to just overcome the force of static friction if the coefficient of friction is 0.309? Round your answer to 2 decimal places.

Respuesta :

LammettHash

Assuming the box is sitting on a flat surface, static friction would have a maximum magnitude of

0.309mg = 0.309 (15.18 kg) (9.80 m/s²) ≈ 45.97 N

so the applied force should have at least this magnitude.

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