A block of mass m is released on a smooth inclined plane of inclination theta with the horizontal. The force exerted by the plane on the block has a magnitude?
Answers
Answered by
62
macos0= mgsin0
a=gtan0
N= mgcos0 + masin0
N= mgcos0 + m(gtan0) sin0
N= mgcos0 + mg (sin0 / cos0) sin0
N= mgcos²0 + mgsin²0/ cos0
= mgcos0(cos²0 + sin²0)
= mg/cos0
a=gtan0
N= mgcos0 + masin0
N= mgcos0 + m(gtan0) sin0
N= mgcos0 + mg (sin0 / cos0) sin0
N= mgcos²0 + mgsin²0/ cos0
= mgcos0(cos²0 + sin²0)
= mg/cos0
Answered by
17
"Ma cos0= mg sin0
a=g tan0
N= mg cos0 + ma sin0
N= mg cos0 + m (gtan0) sin0
N= mg cos0 + mg (sin0 / cos0) sin0
N= mg cos²0 + mgsin²0/ cos0
= mg cos0 (cos²0 + sin²0)
= mg / cos0
Block of mass m is released on the smooth inclined plane with the inclination theta horizontal to force exerted by plane on the block has a magnitude.
"
Similar questions