JEE Main201911 Jan 2019Morning ShiftPhysicsRotational MotionActual
A slab is subjected to two forces F ₁ and F ₂ of same magnitude F as shown in the figure. Force F ₂ is in XY- plane while force F ₁ acts along z -axis at the point (2 i +3 j ) . The moment of these forces about point O will be:
Options
- A(3 i -2 j +3 k ) F
- B(3 i -2 j -3 k ) F
- C(3 i +2 j -3 k ) F
- D(3 i +2 j +3 k ) F
Correct answer
A. (3 i -2 j +3 k ) F
Step-by-step solution
Given, F ₁= F 2 (- i )+ F 3 2 (- j ) r ₁=0 i +6 j Torque due to F ₁ force _ F ₁ = r ₁ F ₁=6 j ( F 2 (- i )+ F 3 2 (- j ) )=3 ~F ( k ) Torque due to F ₂ force aligned _ F ₂ =(2 i +3 j ) & F k =3 ~F i +2 ~F (- j ) _ net = _ F ₁ + _ F ₂ &=3 Fi +2 ~F (- j )+3 ~F ( k ) &=(3 i -2 j +3 k ) F aligned