An equilateral triangular plate OPQ of side 10 cm lies in the xy-plane with O at the origin and Q on the positive x-axis, so that the apex P is at (5, 5√3) cm. A force ⃗F=4̂i-3̂j is applied at P. The torques about O and about Q (in N cm) are, respectively

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(2) -15-20√3, 15-20√3. Work in centimetres, so the torques come out in N cm. O=(0,0), Q=(10,0) and P=(5, 5√3), since the height of an equilateral triangle of side 10 is 5√3. For forces in a plane, the torque about a point is the scalar τ=x F_y - y Fₓ, with (x,y) measured from that point to where the force acts. About O: (x,y)=(5, 5√3), so τ_O=(5)(-3)-(5√3)(4)=-15-20√3. About Q: (x,y)=(5-10, 5√3-0)=(-5, 5√3), so τ_Q=(-5)(-3)-(5√3)(4)=15-20√3. Both are negative, as they must be: the force has a downward pull at a point above both reference points, so it turns the plate clockwise about each. Check: τ_O-τ_Q=-30, which matches (10̂i)×(4̂i-3̂j)=-30̂k.

An equilateral triangular plate OPQ of side 10 cm lies in the xy-plane with O at the origin and Q on the positive x-axis, so that the apex P is at (5, 5√3) cm. A force ⃗F=4̂i-3̂j is applied at P. The torques about O and about Q (in N cm) are, respectively

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