Force F = (-5.3 N) ? + (3.2 N); acts on a particle with position vector...

Question

Question

science physics

Answer 1

Answer #1

Torque= R×F= (2.4 i^ + 3.9 j^)×(-5.3i^ + 3.2j^)= 21.438k^ Nm

a) magnitude= 21.438 Nm

b) we know that |R|•|F| sinø= 21.438

=>√(2.4^2 + 3.9^2) × √(5.3^2 + 3.2^2)×sinø=21.438

Hence sinø= 0.7561

Hence ø=49.1°

Answer 2

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