Ronald who weighs 42 kg likes to slide down the banister of the stairs when he goes down from the second floor of his house to the first floor. If the banister is inclined at an angle of 300 to the horizontal and the coefficient of sliding friction between Ronald and the banister is 0.25, what is the force of friction impeding Ronald’s motion down the banister?
As it's now been over a month since this question (most likely homework, but possibly some other kind of academic effort) was asked, it will give no unfair advantage to the OP to answer it now.
The coefficient of sliding friction is defined as the ratio of the friction force, impeding the movement of an object sliding along a surface, to the normal force of that object on that surface (ie, the force exactly perpendicular to the surface).
I'll assume the banner is at a 30 degree angle, instead of 300. I'll also assume that Ronald is in a place (like the surface of our Earth) where the acceleration of gravity is 9.8 m/sec^2
Simple trigonometry of the above diagram shows that
F(normal) = Cos(ɵ) x Force(gravity)
The gravity force is given by
F(gravity) = mass x gravity acceleration
F(g) = (42 kg) x (9.8 m/sec^2)
= 410 newtons
F(normal) = Cos (30 degrees) x 410 newtons
= (.866) x 410 newtwons
= 356 newtons
F(friction) = (coefficient of sliding friction) x F(normal)
= (.25) x 356 newtons)
= 89 newtons