202301111028 Solution to 2003-IBHL-I-4

Two forces of magnitudes $7\,\mathrm{N}$ and $5\,\mathrm{N}$ act at a point. Which one of the following is not a possible value for the magnitude of the resultant force?

A. $1\,\mathrm{N}$
B. $3\,\mathrm{N}$
C. $5\,\mathrm{N}$
D. $7\,\mathrm{N}$

Roughwork.

Suppose

$\begin{aligned} \mathbf{F}_1 & = 7\sin\alpha\,\hat{\mathbf{i}} + 7\cos\alpha\,\hat{\mathbf{j}} \\ \mathbf{F}_2 & = 5\sin\beta\,\hat{\mathbf{i}} + 5\cos\beta\,\hat{\mathbf{j}} \\ \end{aligned}$

where $\alpha ,\beta\in [0,2\pi )$ .

$\begin{aligned} \mathbf{F}_3 & = \mathbf{F}_1+\mathbf{F}_2 \\ & = (7\sin\alpha + 5\sin\beta )\,\hat{\mathbf{i}} + (7\cos\alpha +5\cos\beta )\,\hat{\mathbf{j}} \\ F_3 & = |\mathbf{F}_3| \\ & = \sqrt{(7\sin\alpha + 5\sin\beta )^2 + (7\cos\alpha +5\cos\beta )^2}\\ & = \sqrt{74+70(\sin\alpha\sin\beta +\cos\alpha\cos\beta )}\\ & = \sqrt{74+70\cos (\alpha-\beta)} \\ \end{aligned}$

$\begin{aligned} \because \quad & \begin{cases} \max (F_3) = \sqrt{74+70(1)} = 12 \\ \min (F_3) = \sqrt{74+70(-1)} = 2 \\ \end{cases} \\ \therefore\quad & F_3 \in [2,12] \\ \end{aligned}$

One can actually do without vector analysis for the sake of a mental exercise.

This problem is not to be attempted.

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