**Problem 015**

Forces F, P, and T are concurrent and acting in the direction as shown in Fig. P-015.

- Find the value of F and α if T = 450 N, P = 250 N, β = 30Â°, and the resultant is 300 N acting up along the y-axis.
- Find the value of F and α if T = 450 N, P = 250 N, β = 30Â° and the resultant is zero.
- Find the value of α and β if T = 450 N, P = 250 N, F = 350 N, and the resultant is zero.

**Solution 015**

**Part a: Unknown force and direction with non-zero resultant**

$R_x = 0$ and $R_y = 300 \, \text{ N}$

$R_x = \Sigma F_x$

$0 = F \cos \alpha + 250 \cos 30^\circ - 450$

$F \cos \alpha = 233.49$

$F = \dfrac{233.49}{\cos \alpha}$

$R_y = \Sigma F_y$

$300 = F \sin \alpha - 250 \sin 30^\circ$

$F \sin \alpha = 425$

$\left( \dfrac{233.49}{\cos \alpha} \right) \sin \alpha = 425$

$\tan \alpha = 1.8202$

$\alpha = 61.22^\circ$ *answer*

$F = \dfrac{233.49}{\cos 61.22^\circ}$

$F = 484.92 \, \text{ N}$ *answer*

**Part b: Unknown force and direction with zero resultant**

$R_x = 0$ and $R_y = 0$

$R_x = \Sigma F_x$

$0 = F \cos \alpha + 250 \cos 30^\circ - 450$

$F \cos \alpha = 233.49$

$F = \dfrac{233.49}{\cos \alpha}$

$R_y = \Sigma F_y$

$0 = F \sin \alpha - 250 \sin 30^\circ$

$F \sin \alpha = 125$

$\left( \dfrac{233.49}{\cos \alpha} \right)\sin \alpha = 125$

$\tan \alpha = 0.5354$

$\alpha = 28.16^\circ$ *answer*

$F = \dfrac{233.49}{\cos 28.16^\circ}$

$F = 264.85 \, \text{ N}$ *answer*

**Part c: Unknown direction of two forces with zero resultant**

$R_x = 0$ and $R_y = 0$

$R_y = \Sigma F_y$

$0 = 350 \sin \alpha - 250 \sin \beta$

$7 \sin \alpha - 5 \sin \alpha = 0$

$7 \sin \alpha = 5 \sin \beta$

$49 \sin^2 \alpha = 25 \sin^2 \beta$ → Equation (1)

$R_x = \Sigma F_x$

$0 = 350 \cos \alpha + 250 \cos \beta - 450$

$7 \cos \alpha + 5 \cos \beta - 9 = 0$

$7 \cos \alpha = 9 - 5 \cos \alpha$

$49 \cos^2 \alpha = (9 - 5 \cos \alpha)^2$

$49 \cos^2 \alpha = 81 - 90 \cos \beta + 25 \cos^2 \beta$ → Equation (2)

Equation (1) + Equation (2)

$49 \sin^2 \alpha + 49 \cos^2 \alpha = 25 \sin^2 \beta + (81 - 90 \cos \beta + 25 \cos^2 \beta)$

$49(\sin^2 \alpha + \cos^2 \alpha) = 25(\sin^2 \beta + \cos^2 \beta) + 81 - 90 \cos \beta$

$49(1) = 25(1) + 81 - 90 \cos \beta$

$90 \cos \beta = 25 + 81 - 49$

$\cos \beta = \frac{57}{90}$

$\beta = 50.70^\circ$ *answer*

From Equation (1)

$49 \sin^2 \alpha = 25 \sin^2 50.70^\circ$

$7 \sin \alpha = 5 \sin 50.70^\circ$

$\sin \alpha = \frac{5}{7} \sin 50.70^\circ$

$\alpha = 33.56^\circ$ *answer*

**Another Solution for Part c**

$250^2 = 350^2 + 450^2 - 2(350)(450)\cos \alpha$

$\cos \alpha = \dfrac{350^2 + 450^2 - 250^2}{2(350)(450)}$

$\alpha = 33.557^\circ$ *answer*

$350^2 = 250^2 + 450^2 - 2(250)(450)\cos \beta$

$\cos \beta = \dfrac{250^2 + 450^2 - 350^2}{2(250)(450)}$

$\beta = 50.704^\circ$ *answer*