Question

Knowing that α= 60° and that boom AC exerts on pin C a force directed along line AC, determine (a)the magnitude of t hat force, (b) the tension in cable BC.Figure:Cable BC is making an angle alpha with fixed support.

Answer 1

Answer:

Tbc = 230.69 N ; Fac = 172.31 N

Explanation:

Sum of forces in y direction:

$T_{BC} * sin (35) = 300*sin (70) + F_{AC}*sin (60) .... Eq 1$

Sum of forces in x direction:

$T_{BC} * cos (35) + F_{AC}*cos (60)= 300*cos (70) .... Eq 1$

Solving Eq 1 and Eq 2 simultaneously:

$T_{BC} = 281.9077862 + \sqrt{3} / 2 * F_{AC}\\\\F_{AC} (1.736868124) = 300*cos (70) - 491.4912266*cos (35)\\\\F_{AC} = - \frac{300}{1.736868124}\\\\F_{AC} = - 172.73 N\\\\T_{BC} = 230.69 N$

Answer: Tbc = 230.69 N ; Fac = 172.31 N

Answer 2

(a) The tension in the cable at AC is -200.67 N.

(b) The  tension in the cable at BC is 328.99 N.

Forces resolved along y-axis

The sum of the forces in y-direction is calculated as follows;

T(BC)sin(35) = 400 x sin(65) + F(AC) sin(60) --- (1)

Forces resolved along x-axis

T(BC)cos(35) + F(AC) cos(60) = 400 x cos(65)  ---(2)

From equation(1);

$T_{BC} = \frac{400 \times sin(65) \ + \ F_{AC} sin(60)}{sin(35)} \\\\T_{BC} = 632 + 1.51F_{AC}$

From equation (2);

0.82T(BC) + 0.5F(AC) = 169.1

$0.82(632 + 1.51F_A_C) + 0.5F_A_C= 169.1\\\\518.24 + 1.24F_A_C + 0.5F_A_C = 169.1\\\\F_A_C = \frac{-349.14}{1.74} \\\\F_A_C = -200.67 \ N$

T(BC) = 632 + 1.51(-200.67)

T(BC) = 328.99 N

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