This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Sample Problem: Friction applied at an angle 1 PHYS 2101, P.M.Shikhaliev LSU, Fall 2010 x y F a x Given: m= 3 kg F = 12 N k = 0.4 Needed: At which angle acceleration will be highest Solution: F net = ma  Newton’s second law F net, x = ma x = F x –f k = Fcos  k F N (1) F net, y = ma y = 0 = F N + F y – F g F N = F g – F y = mg  Fsin (2) Substituting (2) in (1) gives: ma x = Fcos  k (mg  Fsin ) a x = (F/m)cos  k g + ( k F/m)sin a x /d =  (F/m)sin + ( k F/m)cos = 0 sin + k cos = 0 tan = k = 0.4 = 22 o F N F g f k 6.4. The drag force and terminal speed When there is a relative velocity between a fluid and a body (either because the body moves through the fluid or because the fluid moves past the body), the body experiences a drag force, D, that opposes the relative motion and points in the direction in which the fluid flows relative to the body. 2 PHYS 2101, P.M.Shikhaliev LSU, Fall 2010 6.4. Drag force and terminal speed Drag force acts when body moves in fluid or gas. For air, where r is the air density (mass per volume), A is the effective cross sectional area of the body (the area of a cross section taken perpendicular to the velocity), and C is the drag coefficient . Like friction, it is directed against motion C is unitless because: When a body falls through air, its speed increases, and drag force increases too. At speed v t the drag...
View
Full Document
 Spring '07
 GROUPTEST
 Physics, Acceleration, Energy, Force, Friction, Potential Energy, P.M.Shikhaliev LSU

Click to edit the document details