Reactions of Tripod Made from Wood Planks

Plank DEB

$\Sigma M_D = 0$

$4R_B = 2(100 \times 4) + 2R_E$

$2R_B = 400 + R_E$   ←   eq. (1)
 

 

$\Sigma M_B = 0$

$4R_D = 2(100 \times 4) + 4(3500) + 2R_E$

$2R_D = 7400 + R_E$   ←   eq. (2)

 

Plank EFC

$\Sigma M_C = 0$

$4R_E = 2(100 \times 4) + 2R_F$

$2R_E = 400 + R_F$   ←   eq. (3)
 

 

$\Sigma M_E = 0$

$4R_C = 2(100 \times 4) + 2R_F$

$2R_C = 400 + R_F$   ←   eq. (4)

 

Plank FDA

$\Sigma M_A = 0$

$4R_F = 2(100 \times 4) + 2R_D$

$2R_F = 400 + R_D$   ←   eq. (5)
 

 

$\Sigma M_F = 0$

$4R_A = 2(100 \times 4) + 2R_D$

$2R_A = 400 + R_D$   ←   eq. (6)

 

eq. (1) - eq. (2)

$2R_B - 2R_D = -7000$

$R_D = R_B + 3500$
 

From eq. (6)
$2R_A = 400 + (R_B + 3500)$

$2R_A - R_B = 3900$   ←   eq. (7)

 

eq. (3) - eq. (4)

$2R_E - 2R_C = 0$

$R_E = R_C$
 

From eq. (1)
$2R_B = 400 + R_C$

$2R_B - R_C = 400$   ←   eq. (8)

 

eq. (5) - eq. (6)

$2R_F - 2R_A = 0$

$R_F = R_A$
 

From eq. (4)
$2R_C = 400 + R_A$

$-R_A + 2R_C = 400$   ←   eq. (9)

 

Using the Calculator for eq. (7), eq. (8), and eq. (9)

$R_A = 2400 ~ \text{N}$       Answer: [ D ]

$R_B = 900 ~ \text{N}$       Answer: [ A ]

$R_C = 1400 ~ \text{N}$       Answer: [ C ]

$P = 3500 ~ \text{N}$

$W = 3[ \, 100(4) \, ] = 1200 ~ \text{N}$
 

$a = 2 \sin 60^\circ = \sqrt{3} ~ \text{m}$

$b = 2 \cos 60^\circ = 1 ~ \text{m}$
 

$\Sigma M_{BG} = 0$

$bR_A + (4 + b)R_C = 2bP + 2bW$

$1R_A + (4 + 1)R_C = 2(1)(3500) + 2(1)(1200)$

$R_A + 5R_C = 9400$   ←   eq. (1)
 

$\Sigma M_{BI} = 0$

$3aR_A + aR_C = 2aP + (a + \frac{1}{3}a)W$

$3R_A + R_C = 2P + \frac{4}{3}W$

$3R_A + R_C = 2(3500) + \frac{4}{3}(1200)$

$3R_A + R_C = 8600$   ←   eq. (2)
 

From eq. (1) and eq. (2)
$R_A = 2400 ~ \text{N}$

$R_C = 1400 ~ \text{N}$
 

$\Sigma M_{HI} = 0$

$4R_A + (4 + b)R_B = 3P + 3W$

$4R_A + (4 + 1)R_B = 3(3500) + 3(1200)$

$4R_A + 5R_B = 14,100$

$4(2400) + 5R_B = 14,100$

$R_B = 900 ~ \text{N}$
 

Summary

$R_A = 2400 ~ \text{N}$       Answer: [ D ]

$R_B = 900 ~ \text{N}$       Answer: [ A ]

$R_C = 1400 ~ \text{N}$       Answer: [ C ]

$P = 3500 ~ \text{N}$

$W = 3[ \, 100(4) \, ] = 1200 ~ \text{N}$
 

From triangle BEC in the figure below

$x^2 = 2^2 + 4^2 - 2(2)(4) \cos 120^\circ$

$x^2 = 28$

$x = 2\sqrt{7} ~ \text{m}$
 

$\dfrac{x}{\sin 120^\circ} = \dfrac{2}{\sin \alpha}$

$\sin \alpha = \dfrac{2\sin 120^\circ}{x}$

$\sin \alpha = \dfrac{2\sin 120^\circ}{2\sqrt{7}}$

$\sin \alpha = \frac{1}{14}\sqrt{21}$
 

$\dfrac{x}{\sin 120^\circ} = \dfrac{4}{\sin \beta}$

$\sin \beta = \dfrac{4\sin 120^\circ}{x}$

$\sin \beta = \dfrac{4\sin 120^\circ}{2\sqrt{7}}$

$\sin \beta = \frac{1}{7}\sqrt{21}$

 

$a = x \sin 60^\circ = 2\sqrt{7} \sin 60^\circ = \sqrt{21} ~ \text{m}$

$b = 4 \sin \beta = 4(\frac{1}{7}\sqrt{21}) = \frac{4}{7}\sqrt{21} ~ \text{m}$

$c = \frac{1}{3}a = \frac{1}{3}\sqrt{21} ~ \text{m}$

$d = 2 \sin \alpha = 2(\frac{1}{14}\sqrt{21}) = \frac{1}{7}\sqrt{21}$

$e = 4 \sin \alpha = 4(\frac{1}{14}\sqrt{21}) = \frac{2}{7}\sqrt{21}$
 

 

$\Sigma M_{BC} = 0$

$aR_A = bP + cW$

$\sqrt{21}R_A = \frac{4}{7}\sqrt{21}(3500) + \frac{1}{3}\sqrt{21}(1200)$

$R_A = \frac{4}{7}(3500) + (1200)$

$R_A = 2400 ~ \text{N}$
 

$\Sigma M_{AC} = 0$

$aR_B = dP + cW$

$\sqrt{21}R_B = \frac{1}{7}\sqrt{21}(3500) + \frac{1}{3}\sqrt{21}(1200)$

$R_B = \frac{1}{7}(3500) + \frac{1}{3}(1200)$

$R_B = 900 ~ \text{N}$
 

$\Sigma M_{AB} = 0$

$aR_C = eP + cW$

$\sqrt{21}R_C = \frac{2}{7}\sqrt{21}(3500) + \frac{1}{3}\sqrt{21}(1200)$

$R_C = \frac{2}{7}(3500) + \frac{1}{3}(1200)$

$R_C = 1400 ~ \text{N}$
 

Summary

$R_A = 2400 ~ \text{N}$       Answer: [ D ]

$R_B = 900 ~ \text{N}$       Answer: [ A ]

$R_C = 1400 ~ \text{N}$       Answer: [ C ]