physics

thanks! The frictional force that keeps the roofer from slipping isfparallel. The magnitude of this force is equal to the maximum frictional force available for the roofer, which is given by the coefficient of friction multiplied by the normal force: Ffparallel = μN Where μ is the coefficient of friction and N is the normal force (mg, where m is the roofer's mass and g is the acceleration due to gravity). In this case, the normal force is equal to 71 kg × 9.8 m/s² = 697.8 N. Therefore, the magnitude of the frictional force equal to μ × 697.8 N. To calculate μ, we need to use the slope of the roof: given that the height of the roof increases by 1.00 m over a horizontal distance of 3.5 m, the slope of the roof is equal to tan−1(1.00/3.5). Therefore, the coefficient of friction would be equal to μ = tan(tan−1(1.00/3.5)). Plugging the value for μ into the equation, the magnitude of the frictional force would be equal to Ffparallel = μN = tan(tan−1(1.00/3.5)) × 697.8 N ≈ 363.6 N.