Question

a crate weighing 562 newtons is resting on an inclined plane 30 degress above the horizontal. Find the components of the weight forces that are parallel and perpendicular to the plane?
Wc = mg = 562 N. = Weight of crate.

Fc = 562 N. @30 Deg. = Force of crate.
Fp = 562*sin30 = 281 N. = Force parallel to plane.
Fv = 562*cos30 = 487 N. = Force perpendicular to plane.
270-30=240
a crate weighing 562 newtons is resting on an inclined plane 30 degress above the horizontal. Find the components of the weight forces that are parallel and perpendicular to the plane?
Fgx= 562cos(30.0)=486N
Fly= 562sin(30.0)=281
Tan-1(281 divides 486)
yby

Related Questions

A 61-kg person on skis is going down a hill sloped at 30¡Æ from the horizontal. The coefficient of f... An crate weighing 576 N is resting on a plane inclined 29¡Æ above the horizontal. The magnitude of t... An crate weighing 576 N is resting on a plane inclined 31° above the horizontal. (a) Calculate the...