Engineering, 05.10.2020 06:01 moneywaydaedae

Six forces act on a beam that forms part of a building's frame. The vector sum of the forces is zero. The magnitudes

|FB| = Fel 20 kN, Fc = 16 kN, and (Fpl = 9 kN.

Determine the magnitudes of FA and FG.

Answers: 3

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From the given information, we can say that the sum of the upward forces is equivalent to the sum of the downward forces.

From the diagram, equating the component of the upward forces and the downward forces, we have:

$\mathbf{-F_a sin 70 +F_c sin 40+F_d sin 40 + F_g sin50= F_b+F_e}$ --- (1)

Also, the sum of the horizontal positive x-axis as well as the horizontal negative x-axis can be computed as:

$\mathbf{F_g cos 50 +F_d cos 40 = F_c cos 40 +F_a cos 70 --- (2)}$

If:

$F_B = F_E = 20 \ kN \\ \\ F_c = 16 \ kN \\ \\ F_D= 9\ kN$

Then, from equation (1), we can have the following:

$\mathbf{F_a sin 70 + 16 sin 40 + 9 sin40 + F_gsin 50 = (20 + 20 )kN}$

collecting like terms;

$\mathbf{F_a sin 70 + F_gsin 50 = 40 - 16 sin 40 - 9 sin40 }$

$\mathbf{F_a sin 70 + F_gsin 50 =23.93}$ --- (3)

From equation (2);

$\mathbf{F_g cos 50 + 9cos 40 = 16 cos 40 + F_a cos 70}$

collecting like terms:

$\mathbf{F_g cos 50 -F_a cos 70 =16cos 40 -9cos 40}$

$\mathbf{-F_a cos 70 +F_g cos 50 =5.36 ----- (let \ this \ be \ equation (4))}$

Suppose we equate (3) and (4) together using the elimination method;

$\mathbf{F_a sin 70 + F_gsin 50 =23.93}$ --- (3)

$\mathbf{-F_a cos 70 +F_g cos 50 =5.36 --- (4)}$

Let's multiply (3) with ( cos 70 ) and (4) with (sin 70);

Then, we have:

$\mathbf{F_a sin 70 cos 70 + F_gsin 50 cos 70 =23.93 cos 70}$

$\mathbf{-F_a cos 70 sin 70 +F_g cos 50 sin 70 =5.36 sin 70}$

Adding both previous equations together, we have:

$\mathbf{F_a sin 70 cos 70 + F_gsin 50 cos 70 =23.93 cos 70}$

$\mathbf{-F_a cos 70 sin 70 +F_g cos 50 sin 70 =5.36 sin 70}$

$\mathbf{(0 + F_g(sin 50 cos 70 + sin70 cos50) = 23.93 cos 70 + 5.36 sin 70)}$

$\mathbf{( F_g(0.262 + 0.604)) =(8.19 + 5.04)}$

$\mathbf{( F_g(0.866)) =(13.23)}$

$\mathbf{ F_g =\dfrac{(13.23)}{(0.866)}}$

$\mathbf{ F_g =15.28 \ N}$

Replacing the value of $\mathbf{F_g}$ into equation (3), to solve for $\mathbf{F_a}$ , we have:

$\mathbf{F_a sin 70 + F_gsin 50 =23.93}$

$\mathbf{F_a sin 70 + 15.28sin 50 =23.93} \\ \\ \mathbf{F_a sin 70 +11.71 =23.93} \\ \\ \mathbf{F_a sin 70 =23.93-11.71 } \\ \\ \mathbf{F_a sin 70 =12.22 } \\ \\ \mathbf{F_a =\dfrac{12.22 }{sin 70}} \\ \\$

$\mathbf{F_a =13.01 \ N}$

Therefore, we can conclude that the magnitudes of $\mathbf{F_a}$ and $\mathbf{F_g}$ are 13.0 N and 15.28 N respectively.

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Answer from: ericka79

FA = 13 kN FG = 15.3 kN

Explanation:

write each force in terms of magnitude and directions

Fx = F sin Ф

Fy = F cos Ф

where Ф is to be measured from x axis.

∑F at y = o

FAy + FBy + FCy + FDy + FEy + FGy = 0

∑F at x = o

FAx + FBx + FCx + FDx + FEx + FGx = 0

Let

FA = FA sin (110) + FA cos (110)

FB = 20 sin (270) + 20 cos (270)

FC = 16 sin (140) + 16 cos (140)

FD = 9 sin (40) + 9 cos (40)

FE = 20 sin (270) + 20 cos (270)

FG = FG sin (50) + FG cos (50)

add x and y forces:

FAx + FBx + FCx + FDx + FEx + FGx = 0

FAy + FBy + FCy + FDy + FEy + FGy = 0

FA sin (110) + 0 + 16 sin (140) + 9 sin (40) + 0 + FG sin (50) = 0

FA cos (110) - 20 + 16 cos (140) + 9 cos (40) - 20 + FG cos (50 = 0

FA sin (110) + 0 + 10.285 + 5.785 + 0 + FG sin (50) = 0

FA cos (110) - 20 - 12.257 + 6.894 - 20 + FG cos (50) = 0

FA sin (110) + 16.070 + FG sin (50) = 0

FA cos (110) - 45.363 + FG cos (50) = 0

solving for FA, and FG

FA = 13 kN

FG = 15.3 kN

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