The magnitude and direction of the acceleration produced in a body of mass 5 kg when two mutually perpendicular forces 8 N and 6 N act on it, are respectively:

First, resolve the two perpendicular forces into orthogonal components. Let the 8 N force act along the x‑axis and the 6 N force act along the y‑axis.

Calculate the magnitude of the resultant force using the Pythagorean theorem:

$F\_\{\backslash text\{result\}\}\; =\; \backslash sqrt\{8^\{2\}\; +\; 6^\{2\}\}\; =\; \backslash sqrt\{64\; +\; 36\}\; =\; \backslash sqrt\{100\}\; =\; 10\backslash \; \backslash text\{N\}$

Now apply Newton’s second law, \(\mathbf{F}=m\mathbf{a}\), to find the magnitude of the acceleration:

$a\; =\; \backslash frac\{F\_\{\backslash text\{result\}\}\}\{m\}\; =\; \backslash frac\{10\backslash \; \backslash text\{N\}\}\{5\backslash \; \backslash text\{kg\}\}\; =\; 2\backslash \; \backslash text\{m·s\}^\{-2\}$

Next, determine the direction of the acceleration relative to the 8 N force (the x‑axis). The angle \(\theta\) satisfies

$\backslash tan\backslash theta\; =\; \backslash frac\{6\}\{8\}\; =\; \backslash frac\{3\}\{4\}$

Hence

$\backslash theta\; =\; \backslash tan^\{-1\}\backslash !\backslash left(\backslash frac\{3\}\{4\}\backslash right)$

This angle is measured from the direction of the 8 N force toward the 6 N force.

Therefore, the acceleration of the 5 kg body has a magnitude of 2 m·s⁻² and is directed at an angle \(\tan^{-1}(3/4)\) measured from the line of action of the 8 N force.