Three forces \(-63j , 32.14i + 38.3j\) and \(14i - 24.25j\) act on a body of mass 5kg. Find, correct to one decimal place, the :
(a) magnitude of the resultant force ;
(b) acceleration of the body.
Add the forces component by component. The three forces are
\[F_1 = -63\mathbf{j},\quad F_2 = 32.14\mathbf{i} + 38.3\mathbf{j},\quad F_3 = 14\mathbf{i} - 24.25\mathbf{j}.\]
Resultant components.
\[\sum F_x = 0 + 32.14 + 14 = 46.14,\qquad \sum F_y = -63 + 38.3 - 24.25 = -48.95.\]
So \(\mathbf{R} = 46.14\mathbf{i} - 48.95\mathbf{j}\).
(a) Magnitude of the resultant force.
\[|\mathbf{R}| = \sqrt{46.14^2 + 48.95^2} = \sqrt{2128.90 + 2396.10} = \sqrt{4525.00} \approx 67.3\ \text{N}.\]
(b) Acceleration. By Newton's second law \(\mathbf{R} = m\mathbf{a}\), with \(m = 5\ \text{kg}\):
\[a = \frac{|\mathbf{R}|}{m} = \frac{67.3}{5} \approx 13.5\ \text{m s}^{-2}.\]
The acceleration is directed along the resultant force, with magnitude \(13.5\ \text{m s}^{-2}\).