uhh, we haven't discussed anything about this.. i dont know how to do this
What is ask in the first drawing?
For the second drawing: $M_z = 0$ ← Force F passes through the z-axis
$d = \sqrt{4^2 + 6^2 + 8^2}$
$d = 2\sqrt{29} ~ \text{m}$
$F_x = 230 \times \dfrac{4}{2\sqrt{29}} = 85.42 ~ \text{kN}$
$F_y = 230 \times \dfrac{6}{2\sqrt{29}} = 128.13 ~ \text{kN}$
$M_x = 8F_y = 1025.04 ~ \text{N}\cdot\text{m}$
$M_y = 8F_x = 683.36 ~ \text{N}\cdot\text{m}$
the first question sir is, "find the centroid of the shaded figure below"
sir thank you for your response, i'll study it
By symmetry $\bar{x} = 10$
Equation of L1: $\dfrac{x}{19} + \dfrac{y}{21} = 1$
$y = 21\left( 1 - \dfrac{x}{19} \right)$
At x = 10 $y = 21\left( 1 - \dfrac{10}{19} \right)$
$y = \frac{180}{19}$
$A_1 = 20(21) = 420$
$A_2 = 2 \times \frac{1}{2}(1)(21) = 21$
$A_3 = \frac{1}{2}(20)(21 - y) = \frac{2190}{19}$
$A_4 = \frac{1}{2}(18y) = \frac{1620}{19}$
$A = A_1 - A_2 - A_3 - A_4$
$y_1 = \frac{21}{2}$
$y_2 = \frac{21}{3}$
$y_3 = 21 - \frac{1}{3}(21 - y)$
$y_4 = \frac{1}{3}y$
$A\bar{y} = A_1 y_1 - A_2 y_2 - A_3 y_3 - A_4 y_4$ and solve for $\bar{y}$.
Centroid: $(10, \bar{y})$
sir thank you sir! i understand it pretty well. (maybe hehe)
*correction sir y=21(1−10/19) y=189/19 *and sir about this: x/19+y/21=1, why does it have to be equal to 1?
SIR I REALLY REALLY APPRECIATE THIS SITE AND YOUR HELP THANK YOU SO MUCH
i understand now sir, it happens that my first try was kinda right.. 10/x = 9/y = 19/21 but the answer is quite unbelievable.. centroid is (10,19.35) O.O
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What is ask in the first drawing?
For the second drawing:
$M_z = 0$ ← Force F passes through the z-axis
$d = \sqrt{4^2 + 6^2 + 8^2}$
$d = 2\sqrt{29} ~ \text{m}$
$F_x = 230 \times \dfrac{4}{2\sqrt{29}} = 85.42 ~ \text{kN}$
$F_y = 230 \times \dfrac{6}{2\sqrt{29}} = 128.13 ~ \text{kN}$
$M_x = 8F_y = 1025.04 ~ \text{N}\cdot\text{m}$
$M_y = 8F_x = 683.36 ~ \text{N}\cdot\text{m}$
the first question sir is, "find the centroid of the shaded figure below"
sir thank you for your response, i'll study it
By symmetry
$\bar{x} = 10$
Equation of L1:
$\dfrac{x}{19} + \dfrac{y}{21} = 1$
$y = 21\left( 1 - \dfrac{x}{19} \right)$
At x = 10
$y = 21\left( 1 - \dfrac{10}{19} \right)$
$y = \frac{180}{19}$
$A_1 = 20(21) = 420$
$A_2 = 2 \times \frac{1}{2}(1)(21) = 21$
$A_3 = \frac{1}{2}(20)(21 - y) = \frac{2190}{19}$
$A_4 = \frac{1}{2}(18y) = \frac{1620}{19}$
$A = A_1 - A_2 - A_3 - A_4$
$y_1 = \frac{21}{2}$
$y_2 = \frac{21}{3}$
$y_3 = 21 - \frac{1}{3}(21 - y)$
$y_4 = \frac{1}{3}y$
$A\bar{y} = A_1 y_1 - A_2 y_2 - A_3 y_3 - A_4 y_4$ and solve for $\bar{y}$.
Centroid: $(10, \bar{y})$
sir thank you sir! i understand it pretty well. (maybe hehe)
*correction sir y=21(1−10/19)
y=189/19
*and sir about this:
x/19+y/21=1, why does it have to be equal to 1?
SIR I REALLY REALLY APPRECIATE THIS SITE AND YOUR HELP THANK YOU SO MUCH
i understand now sir, it happens that my first try was kinda right.. 10/x = 9/y = 19/21
but the answer is quite unbelievable.. centroid is (10,19.35) O.O
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