Question

Force F = (2.0 N) i - (3.0 N) k acts on a pebble with positionvector r = (O.50 m) j - (2.0 m) k, relative to the origin. What isthe resulting torque acting on the pebble about a.) the origin and b.) a point with coordinates (2.0 m, 0, - 3.0 m)?

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asked 2021-03-07
Force F = (2.0 N) i - (3.0 N) k acts on a pebble with positionvector r = (O.50 m) j - (2.0 m) k, relative to the origin. What isthe resulting torque acting on the pebble about
a.) the origin and
b.) a point with coordinates (2.0 m, 0, - 3.0 m)?

Expert Answers (1)

2021-03-08
The Force F = 2 i - 3 k the radius vectir r = 0.50 i -2.0 k where i , j and k are the unit vectors in the x, y and zdirection.
The Torque
t= r xF = ( 0.50 j -2.0 k ) x ( 2 i - 3 k )
t= - 2m i + 0.5m j +1.0m k
b) The torque about the point (2.0 m, 0, - 3.0 m) is
t' = r ' x Fr ' = r - ( 2i - 3k ) = ( 0.50 j -2.0 k )- ( 2i - 3k )
0r r ' = - 2 i + 0.5 j + k
Therefore
t' = ( -2 i + 0.5 j + k ) x ( 2i - 3k )
0rt' = - 1.5m i- 4m j -1.0m k
Kindly confirm whether my answer iscorrect. Thanks.
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