Question

The force vector F\mathbf{F} points along the straight line from point AA to point BB. Use Eqs. (2.28)-(2.31) to prove that

rB×F=rA×F.\mathbf{r}_B \times \mathbf{F}=\mathbf{r}_A \times \mathbf{F} .

Strategy: Let rAB\mathbf{r}_{A B} be the position vector from point AA to point B. Express rB\mathbf{r}_B in terms of rA\mathbf{r}_A and rAB\mathbf{r}_{A B}. Notice that the vectors rAB\mathbf{r}_{A B} and F\mathbf{F} are parallel.

Solution

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Let rABr_{AB} the position vector from point A to B.

Express rBr_{B} in terms of rA and rAB.r_{A} \ and \ r_{AB}.

We have:

rB=rA+rABr_{B}=r_{A} + r_{AB}

Therefore,

rB×F=(rA+rAB)×F=(rA×F)+(rAB×F)r_{B} \times F = (r_{A} + r_{AB}) \times F = (r_{A} \times F)+(r_{AB} \times F)

The last term is zero since rABr_{AB} is in parallel with F

Therefore,

rB×F=rA×F\boxed{r_{B} \times F = r_{A} \times F}

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