# 002 Components of forces with given slope

**Problem 002**

Compute the x and y components of each of the four forces shown in Fig. P-002.

**Solution 002**

$P_y = 722(\frac{3}{\sqrt{13}}) = 600.74 \, \text{ lb}$

$Q_x = -200 \cos 60^\circ = -100 \, \text{ lb}$

$Q_y = 200 \sin 60^\circ = 173.20 \, \text{ lb}$

$F_x = -448(\frac{2}{\sqrt{5}}) = -400.70 \, \text{ lb}$

$F_y = -448(\frac{1}{\sqrt{5}}) = -200.35 \, \text{ lb}$

$T_x = 400 \sin 20^\circ = 136.81 \, \text{ lb}$

$T_y = -400 \cos 20^\circ = -375.88 \, \text{ lb}$

**Rectangular Representation**

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

${\bf P} = 722 (\frac{2}{\sqrt{13}} {\bf i} + \frac{3}{\sqrt{13}} {\bf j}) = 400.49 {\bf i} + 600.74 {\bf j} \, \text{ lb}$

${\bf Q} = 200 (-\cos 60^\circ {\bf i} + \sin 60^\circ {\bf j}) = -100 {\bf i} + 173.20 {\bf j} \, \text{ lb}$

${\bf F} = 448 (-\frac{2}{\sqrt{5}} {\bf i} - \frac{1}{\sqrt{5}} {\bf j}) = -400.70 {\bf i} - 200.35 {\bf j} \, \text{ lb}$

${\bf T} = 400 (\sin 20^\circ {\bf i} - \cos 20^\circ {\bf j}) = 136.81 {\bf i} - 375.88 {\bf j} \, \text{ lb}$

The coefficients of **i** and **j** from the vector notations are the respective *x* and *y* components of each force.

**Calculator Techniques**

**Using CMPLX mode: [MODE] → 2:CMPLX**

Using this mode, you input F ∠ θ and the calculator will return *a* + *bi*. The *x*-component of the force is *a* and the *y*-component is *b*. Note that θ starts from the positive side of *x*-axis and is positive if rotated counterclockwise and negative if rotated clockwise.

^{-1}(3/2) = 400.49 + 600.74

**i**

Q = 200 ∠ (180 - 60) = -100 + 173.20**i**

F = 448 ∠ (180 + tan^{-1}(1/2) = -400.70 - 200.35**i**

T = 400 ∠ -(90-20) = 136.81 - 375.88**i**

**Using Rec: [MODE] → 1:COMP**

In this function, we will input Rec(F,θ).

^{-1}(3/2)) =

X = 400.49, Y = 600.74

Q = Rec(200,180-60) =

X = -100, Y = 173.20

F = Rec(448,180+tan^{-1}(1/2)) =

X = -400.70, Y = - 200.35

T = Rec(400,-(90-20)) =

X = 136.81, Y = - 375.88