001 Horizontal and vertical componets of planar forces

Tags:

Problem 001
Problem Determine the x and y components of the forces shown below in Fig P-001.

Solution 001

Click here to show or hide the solution

$F_{x1} = 58 \cos 30^\circ = 50.23 \, \text{ kN}$

$F_{y1} = 58 \sin 30^\circ = 29 \, \text{ kN}$

$F_{x2} = -50 \cos 45^\circ = -35.36 \, \text{ kN}$

$F_{y2} = 50 \sin 45^\circ = 35.36 \, \text{ kN}$

$F_{x3} = -45(\frac{5}{13}) = -17.31 \, \text{ kN}$

$F_{y3} = -45(\frac{12}{13}) = -41.54 \, \text{ kN}$

$F_{x4} = 40 \, \text{ kN}$

$F_{y4} = 0$

Rectangular Representation
${\bf F} = F (\cos \theta_x {\bf i} + \sin \theta_x {\bf j})$

${\bf F_1} = 58 (\cos 30^\circ {\bf i} + \sin 30^\circ {\bf j}) = 50.23 {\bf i} + 29 {\bf j} \, \text{ kN}$

${\bf F_2} = 50 (-\cos 45^\circ {\bf i} + \sin 45^\circ {\bf j}) = -35.36 {\bf i} + 35.36 {\bf j} \, \text{ kN}$

${\bf F_3} = 45 (-\frac{5}{13} {\bf i} - \frac{12}{13} {\bf j}) = -17.31 {\bf i} - 41.54 {\bf j} \, \text{ kN}$

${\bf F_4} = 40 {\bf i} \, \text{ kN}$

From the above vector notation of forces, F_x is the coefficient of i and F_y is the coefficient of j.

Submitted by Romel Verterra on March 18, 2010 - 11:15pm

Recent Updates