- #36
Ukitake Jyuushirou
- 124
- 0
now i have to solve for the coefficient of kinetic friction. kinetic friction force is supposed to be proportional to the Fn. but from wat is given, how do i work out the friction force?
Good. But just for the record, here's how I would do it:Ukitake Jyuushirou said:w=mg
196=20x9.8
Fn = 196 (no vertical force)
196-40 = 156 (force of 40N in +y-component)
Fn = 156
using ur formulaDoc Al said:Good. But just for the record, here's how I would do it:
(Sum of the forces in the vertical direction) = +Fn -mg +80sin(30) = 0
That's the equation for vertical equilibrium. Solving it (trivial, yes) gives:
Fn = mg - 80sin(30)
Now do the same for horizontal forces.
yes ur right. i'd try to use this approach for my other qnsDoc Al said:OK. Again, the way I would do it is:
(Sum of the forces in the horizontal direction) = -Ff +80cos(30) = 0
Solving gives: Ff = 80cos(30)
(The reason for sticking to a systematic approach is: (1) It drives home the key physics principles & (2) It allows you to solve much harder problems.)
Now you can apply the other relationship you know about kinetic friction to solve for [itex]\mu_k[/itex] systematically.