For a 10kg object on a 30degree inclined plane, what are the object's weight and the normal force exerted?

Question

For a 10kg object on a 30degree inclined plane, what are the object's weight and the normal force exerted?

Answer 1

Normal Force = m*g * Cos(30)  where m is 10kg (the object's weight) and g is 9.8 meters/sec

Answer 2

weight=mg=98 N

normal force=98cos30=49sqrt(2) N

accel force=98sin30=49 N

Answer 3

The weight is always equal to mg, so:

w = mg = (10kg)(9.8m/s^2) = 98N

normal force: 98cos30... To verify this, draw a free body diagram. It needs a little knowledge of geometry.

Answer 4

see my answer to your last question

weight is a force due to the acceleration due to gravity (near the earths surface)

F=dp.dt = ma (Newtons 2nd law)

a = g (near earths surface

F=ma=mg

Answer 5

m=10kg,theta=30,w=?,f=?

W=mg=10*10=100N THEREFORE OBJECT weight =100N

F=mgsintheta=10*10*sin30=100*0.5=50N(n... force =50N