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A box of mass 22 kg is at rest on a flat floor. The coefficient of static friction between the box and floor is 0.37. What is the minimum force needed to set the box in motion across the floor?

Respuesta :

The answer is 80 N because

Force of friction = friction coeff. times weight in newtons

weight = 9.81 (on earth) times mass

.37*9.81*22 = 79.8534

round up to 80

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