Answer:
0.0984
Explanation:
From the first diagram attached below; a free flow diagram shows the interpretation of this question which will be used  to solve this question.
From the diagram, the horizontal component of the force is:
[tex]F_X = F_{cos \ \theta}[/tex]
Replacing 42°  for θ and 87.0° for [tex]F[/tex]
[tex]F_X =87.0 \ N \ *cos \ 42 ^\circ[/tex]
[tex]F_X =64.65 \ N[/tex]
On the other hand, the vertical component  is ;
[tex]F_Y = Fsin \ \theta[/tex]
Replacing 42°  for θ and 87.0° for [tex]F[/tex]
[tex]F_Y =87.0 \ N \ *sin \ 42 ^\circ[/tex]
[tex]F_Y =58.21 \ N[/tex]
However, resolving the vector, let A be the be the component of the mutually perpendicular directions.
The magnitude of the two components is shown in the second attached diagram below and is now be written as A cos θ and A sin θ
The expression for the frictional force is expressed as follows:
[tex]f = \mu \ N[/tex]
Where;
[tex]\mu[/tex] is said to be the coefficient of the friction
N = the  normal force
Similarly the normal reaction (N) = mg - F sin θ
Replacing [tex]F_Y \ for \ F_{sin \ \theta}[/tex]. The normal reaction can now be:
[tex]N = mg \ - \ F_Y[/tex]
By balancing the forces, the horizontal component of the force equals to frictional force.
The horizontal component of the force is given as follows:
[tex]F_X = \mu \ ( mg - \ F_Y)[/tex]
Making [tex]\mu[/tex] the subject of the formular in the above equation; we have the following:
[tex]\mu \ = \ \frac{F_X}{mg - F_Y}[/tex]
Replacing the following values: i.e
[tex]F_X \ = \ 64.65 \ N[/tex]
m = 73 Kh
g  = 9.8 m/s²
[tex]F_Y = \ 58.21 N[/tex]
Then:
[tex]\mu \ = \ \frac{64.65 N}{(73.0 kg)(9.8m/s^2) - (58.21 \ N)}[/tex]
[tex]\mu = 0.0984[/tex]
Thus, the coefficient of friction is = 0.0984
