You pull with a force of 77 n on a piece of luggage of mass 23 kg, but it does
not move. if the coefficient of friction between the luggage and the floor is
0.8, what is the force of static friction acting on the luggage?

The force of static friction acting on the luggage is, Fₓ = 180.32 N

Explanation:

Given data,

The mass of the luggage, m = 23 kg

You pulled the luggage with a force of, F = 77 N

The coefficient of static friction of luggage and floor, μₓ = 0.8

The formula for static frictional force is,

                                      Fₓ = μₓ · η

Where,

                                  η - normal force acting on the luggage 'mg'

Substituting the values in the above equation,

                                   Fₓ = 0.8 x 23 x 9.8

                                        = 180.32 N

Hence, the minimum force require to pull the luggage is, Fₓ = 180.32 N

184N

Explanation:

Frictional force is defined as a force of opposition. This force opposes the moving force. The body placed on a plane surface has a frictional force proportional to the normal reaction force acting on the body.

Mathematically, Ff = nR where;

Ff is the frictional force

n is the coefficient of static friction between the luggage and the floor.

R is the normal reaction which is equal to the weight (W) of the body.

W = mg

Weight = 23×10 = 230N

Since weight = normal reaction

Normal reaction = 230N

coefficient of friction between the luggage and the floor = 0.8

Frictional force (Ff) = 0.8×230

Frictional force = 184N

Therefore, the force of static friction acting on the luggage is 184N

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