Question

Find the resultant of the following forces by component method F1=12 N, South. F2=24N,30 degree north of west F3=15N,75degree South of West and F4=32N,50 degree south of East.

Answer 1

Answer:

East of North is better described as East “from” North, because it means that the angle is measured. Eastward from the Northward direction. This means that the eight ways to describe the direction of an angle.

Answer 2

Given:

F1=12 N South.

F2=24N,30 degree north of west

F3=15N,75degree South of West and

F4=32N,50 degree south of East

Find:

resultant = ?

Solution:

$\vec R = \left(\Sigma_{i=1}^{m} x_{i}\right)\,\hat{i}+\left(\Sigma_{i=1}^{m} y_{i}\right)\,\hat{j}$

Where:

$\ \vec R\ – Resultant,\ measured\ in\ newtons.x_{i} – i\-th x-Component, measured\ in \ newtons.y_{i} – i-th y-Component, measured\ in\ newtons.We describe each known vector below:}$

$\|\vec F_{1}\| = 12\,N, South:\vec F_{1} = -12\,\hat{i}\,\,[N]\|\vec F_{2}\| = 24\,N, \angle = 30^{\circ} North \ of \ west:$

$\vec F_{2} = 24\cdot (-\cos 30^{\circ}\,\hat{i}+\sin 30^{\circ}\,\hat{j})\,[N]\\\\\vec F_{2} = -20.785\,\hat{i}+12\,\hat{j}\,\,[N]\|\vec F_{3}\| = 15\,N, \\\\\angle = 75^{\circ} South of west:$

$\vec F_{3} = 15\cdot (-\cos 75^{\circ}\,\hat{i}-\sin 75^{\circ}\,\hat{j})\,\,[N]\vec F_{3} = -3.883\,\hat{i}-14.489\,\hat{j}\,\,[N]\\\|\vec F_{4}\| = 32\,N, \angle = 50^{\circ} South \ of \ east:\\\vec F_{4} = 32\cdot (\cos 50^{\circ}\,\hat{i}-\sin 50^{\circ}\,\hat{j})\,\,[N]\vec F_{4} = 20.569\,\hat{i} -24.513\,\hat{j}\,\,[N]$

We find the resultant by vectorial sum:

$\vec R = \vec F_{1}+\vec F_{2}+\vec F_{3}+\vec F_{4}\\\vec R = (-12-20.785-3.883+20.569)\,\hat{i}+(12-14.489-24.513)\,\hat{j}\,\,[N]\\\vec R = -16.099\,\hat{i} -27.002\,\hat{j}\,\,[N]$

Hence the resultant of given forces is  $\vec R = -16.099\,\hat{i} -27.002\,\hat{j}\,\,[N].$