Force F = (-8.0 N)i + (6.0 N)j acts on a particle with position vector r = (3.0

Question

Force F = (-8.0 N)i + (6.0 N)j acts on a particle with position vector r = (3.0 m)i + (4.0 m)j. What are (a) the torque on the particle about the origin and (b) the angle between the directions of r and F ? Force F = (-8.0 N)i + (6.0 N)j acts on a particle with position vector r = (3.0 m)i + (4.0 m)j. What are (a) the torque on the particle about the origin and (b) the angle between the directions of r and F ? Force F = (-8.0 N)i + (6.0 N)j acts on a particle with position vector r = (3.0 m)i + (4.0 m)j. What are (a) the torque on the particle about the origin and (b) the angle between the directions of r and F ?

Explanation / Answer

(a)
F =  (-8.0 N) i + (6.0 N) j
r =  (3.0 m) i + (4.0 m) j
Torque = r x F
t = [(3.0 m) i + (4.0 m) j] X [ (-8.0 N) i + (6.0 N) j ]
t = 18.0 k^ + 32 k^
t = 50 k^ Nm

(b)
Angle betwen r & F , = 90o (They are perpendicular)

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