1. Statement of the problem. Problem: A body of weight 18 gm.wt is placed on a rough horizontal plane and a horizontal force of magnitude F acts on it to make it about to move. The angle between the resultant reaction force and the limiting static friction force is $\theta$ where $\cos\theta=\frac{4}{5}$. Find the magnitude of the resultant reaction force. 2. Formula used and important rules. When the body is on the verge of moving, the limiting friction $f$ acts horizontally and the normal reaction $N$ acts vertically. The resultant reaction $R$ is the vector sum of $f$ and $N$, so by Pythagoras we have $$R^2=f^2+N^2$$ The angle $\theta$ between $R$ and the friction $f$ satisfies $$\cos\theta=\frac{f}{R}$$ Also, because the applied force is horizontal, the normal reaction equals the weight, so $N=18$ (gm.wt.). 3. Intermediate work and algebraic manipulation. From $\cos\theta=\frac{f}{R}$ we get $$R=\frac{f}{\cos\theta}$$ Squaring both sides and substituting into the Pythagorean relation gives $$\frac{f^2}{\cos^2\theta}=f^2+N^2$$ Rearrange the equation: $$f^2\left(\frac{1}{\cos^2\theta}-1\right)=N^2$$ Use the trigonometric identity $\frac{1}{\cos^2\theta}-1=\tan^2\theta$ to obtain $$f^2\tan^2\theta=N^2$$ Taking positive square roots (all quantities here are positive) gives $$f\tan\theta=N$$ Divide both sides by $\tan\theta$ and show the cancellation: $$\frac{f\tan\theta}{\tan\theta}=\frac{N}{\tan\theta}\Rightarrow\frac{\cancel{\tan\theta}f}{\cancel{\tan\theta}}=\frac{N}{\tan\theta}\Rightarrow f=\frac{N}{\tan\theta}$$ Now substitute $f$ into $R=\frac{f}{\cos\theta}$ to get $$R=\frac{N}{\tan\theta}\cdot\frac{1}{\cos\theta}=N\cdot\frac{1}{\sin\theta}=N\csc\theta$$ 4. Numerical evaluation and final answer. Given $\cos\theta=\frac{4}{5}$ we have $\sin\theta=\frac{3}{5}$. Therefore $$R=N\csc\theta=18\cdot\frac{5}{3}$$ Simplify the arithmetic and show cancellation: $$R=18\cdot\frac{5}{3}=\frac{\cancel{18}}{\cancel{3}}\cdot5=6\cdot5=30$$ Hence the magnitude of the resultant reaction force is 30 gm.wt.