The force of friction acting on Tyler is given by:
[tex]F=\mu N[/tex] (1)
where [tex]\mu[/tex] is the coefficient of friction, and N is the normal force exerted by the floor on Tyler. However, the normal force is equal to Tyler's weight:
[tex]N=mg[/tex]
where m is Tyler's mass and g is the gravitational acceleration. Therefore, the frictional force (1) becomes
[tex]F=\mu mg=(0.3 )(67 kg)(9.81 m/s^2)=197 N[/tex]
