Question 2.5: The link in Fig. 2-18a is subjected to two forces F1 and F2.......
The link in Fig. 2-18a is subjected to two forces F_1 and F_2. Determine the magnitude and direction of the resultant force.
Scalar Notation. First we resolve each force into its x and y components, Fig. 2-185, then we sum these components algebraically.
\overset{+}→ (F_R)_x = ∑F_x;\quad \quad (F_R)_x = 600 cos 30° N – 400 sin 45° N
= 236.8 N →
+↑(F_R)_y=∑F_y;\quad \quad (F_R)_y = 600 sin 30° N + 400 cos 45° N
= 582.8 N ↑
The resultant force, shown in Fig. 2-18c, has a magnitude of
F_R=\sqrt{(236.8\ N)^2+ (582.8\ N)^2}
= 629 N
From the vector addition,
\theta =tan^{-1}\left(\frac{582.8\ N}{236.8\ N} \right) =67.9^\circ
SOLUTION II
Cartesian Vector Notation. From Fig. 2-18b, each force is first expressed as a Cartesian vector.
F_1 = { 600 cos 30°i + 600 sin 30°j } N
F_2 = { -400 sin 45°i + 400 cos 45°j } N
Then,
F_R = F_1 + F_2 = (600 cos 30° N – 400 sin 45°N)i
+ (600 sin 30° N + 400 cos 45° N)j
= { 236.8i + 582.8j } N
The magnitude and direction of FR are determined in the same manner as before.
NOTE: Comparing the two methods of solution, notice that the use of scalar notation is more efficient since the components can be found directly, without first having to express each force as a Cartesian vector before adding the components. Later, however, we will show that Cartesian vector analysis is very beneficial for solving three-dimensional problems.
