luggage of mass 23 kg is pulled through an airport at a constant velocity. The force applied through the handle is 75N at an angle of 60 degrees. what is the coefficient of friction?

Question

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Answer 1

The horizontal component of the applied force is given by F = 75 N x cos60 = 37.5 N.

As the luggage is traveling at a constant velocity, the force of friction is equal and opposite to the applied force, so we can equate the frictional force to the applied force 37.5 N = μ x N where μ is the coefficient of kinetic friction and N the normal force.

This leads to 37.5 N = μ x m x g = μ x 23 kg x 9.8 m/s². Solve for μ to get

μ = 37.5 N / (23kg x 9.8 m/s²) = 1.7