A particle moves through an xyz coordinate system while a force acts on the particle. When the particle has the position vector r = (2.00 m) - (3.00 m) + (2.00 m) , the force is F = Fx + (7.00 N) - (6.00 N) , and the corresponding torque about the origin is T = (4.00 N·m) + (10.0 N·m) + (11.0 N·m) . Determine Fx.
here,
the position vector , r = ( 2 i - 3 j + 2 k) m
the applied force , F = ( Fx i - 6 j) N
the torque applied , T = r X F
(4 i + 10 j + 11 k) N.m = (2i - 3j + 2 k) X ( Fx i + 7 j - 6 k)
(4 i + 10 j + 11 k) N.m = (4 i + (2 Fx + 12) j + (14 + 3 Fx) k)
on compairing
we get
Fx = - 1 N
the value of Fx is (-1 N)
A particle moves through an xyz coordinate system while a force acts on the particle. When...
A particle moves through an xyz coordinate system while a force acts on the particle. When the particle has the position vector r Overscript right-arrow EndScripts = (2.00 m) i Overscript ̂ EndScripts - (3.00 m) j Overscript ̂ EndScripts + (2.00 m) k Overscript ̂ EndScripts, the force is Upper F Overscript right-arrow EndScripts = Fx i Overscript ̂ EndScripts + (7.00 N) j Overscript ̂ EndScripts - (6.00 N) k Overscript ̂ EndScripts, and the corresponding torque about...
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