Find the magnitude and direction of a reaction force on a wall

Key takeaway: This is a standard vector-mechanics question: once the force is resolved into components, the magnitude and the angle both fall out cleanly.

Let the force be resolved into components parallel and perpendicular to the wall.

• If the horizontal component is $R_x$ and the vertical component is $R_y$, then the magnitude is $|R|=\sqrt{R_x^2+R_y^2}.$
• The acute angle $\theta$ it makes with the wall is found using the component along the wall and the component normal to it: $\tan\theta=\frac{\text{perpendicular component}}{\text{parallel component}}.$

So the working is always:

1. identify the two orthogonal components from the diagram or earlier parts,
2. use Pythagoras for the magnitude,
3. use tangent for the angle.

If the force is already written in vector form, say $\mathbf{R}=ai+bj$, then $|R|=\sqrt{a^2+b^2}$ and the angle to the wall is the angle between $\mathbf{R}$ and the wall direction, obtained from the relevant component ratio.

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Pitfalls the pros know 👇 A common slip is measuring the angle from the wrong reference line. The question asks for the angle with the wall, not with the floor or with the normal. Also, keep the acute angle only; if your calculator gives an obtuse result, convert it to the acute one that matches the diagram.

What if the problem changes? If the force is given by $\mathbf{R}=ai+bj$ and the wall is vertical, then the angle with the wall is measured from the $j$-axis. In that case, $\cos\theta=\frac{|b|}{\sqrt{a^2+b^2}}.$ If the wall were horizontal instead, you would swap the roles of $a$ and $b$ in the angle formula.

Tags: resultant force, vector components, angle to a line