Problem 603 | Resultant of Concurrent Forces in Space | Engineering Mechanics Review at MATHalino

Problem 603
Determine the magnitude of the resultant, its pointing, and its direction cosines for the following system of non-coplanar concurrent forces. 100 lb (2, 3, 4); 300 lb (–3, –4, 5); 200 lb, (0, 0, 4).

Solution 603

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Distance

$d = \sqrt{x^2 + y^2 + z^2}$

Components of given forces

From

$\dfrac{F_x}{x} = \dfrac{F_y}{y} = \dfrac{F_z}{z} = \dfrac{F}{d}$

$F_x = \dfrac{x \, F_x}{d}$

$F_y = \dfrac{y \, F_y}{d}$

$F_z = \dfrac{z \, F_z}{d}$

Components of resultant

$R_x = \Sigma F_x$

$R_y = \Sigma F_y$

$R_z = \Sigma F_z$

Resultant

$R = \sqrt{{R_x}^2 + {R_y}^2 + {R_z}^2}$

Force F	Components of Distance			Distance d	Components of Force
Force F	x	y	z	Distance d	F_x	F_y	F_z
100	2	3	4	5.385	37.139	55.709	74.278
300	–3	–4	5	7.071	–127.279	–169.706	212.132
200	0	0	4	4.000	0.000	0.000	200.000
SUM (R_x, R_y, and R_z)					-90.140	–113.997	486.410
RESULTANT					R = 507.657 lb

Direction cosines of the resultant

$\cos \theta_x = \dfrac{R_x}{R} = -0.1776$

$\cos \theta_y = \dfrac{R_y}{R} = -0.2246$

$\cos \theta_z = \dfrac{R_z}{R} = 0.9581$

Another Solution (Using Vectors)

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Direction vectors

${\bf r}_1 = 2{\bf i} + 3{\bf j} + 4{\bf k}$

${\bf r}_2 = -3{\bf i} - 4{\bf j} + 5{\bf k}$

${\bf r}_3 = 0{\bf i} + 0{\bf j} - 3{\bf k}$

Unit vectors

${\bf \lambda}_1 = \dfrac{ 2{\bf i} + 3{\bf j} + 4{\bf k}}{\sqrt{2^2 + 3^2 + 4^2}} = 0.3714{\bf i} + 0.5571{\bf j} + 0.7428{\bf k}$

${\bf \lambda}_2 = \dfrac{-3{\bf i} - 4{\bf j} + 5{\bf k}}{\sqrt{3^2 + 4^2 + 5^2}} = -0.4243{\bf i} - 0.5657{\bf j} + 0.7071{\bf k}$

${\bf \lambda}_3 = \dfrac{0{\bf i} + 0{\bf j} + 4{\bf k}}{\sqrt{0^2 + 0^2 + 3^2}} = 0{\bf i} + 0{\bf j} + 1{\bf k}$

Forces in rectangular form

${\bf F} = F{\bf \lambda}$

${\bf F}_1 = 100(0.3714{\bf i} + 0.5571{\bf j} + 0.7428{\bf k}) = 37.14{\bf i} + 55.71{\bf j} + 74.28{\bf k} ~ \text{lb}$

${\bf F}_2 = 300(-0.4243{\bf i} - 0.5657{\bf j} + 0.7071{\bf k}) = -127.29{\bf i} - 169.71{\bf j} + 212.13{\bf k} ~ \text{lb}$

${\bf F}_3 = 200(0{\bf i} + 0{\bf j} + 1{\bf k}) = 0{\bf i} + 0{\bf j} + 200{\bf k} ~ \text{lb}$

Resultant

${\bf R} = {\bf F}_1 + {\bf F}_2 + {\bf F}_3$

Sum up the coefficients of i, j and k
${\bf R} = -90.15{\bf i} - 114{\bf j} + 486.41{\bf k} ~ \text{lb}$ answer

Magnitude of resultant

$R = \sqrt{90.15^2 + 114^2 + 486.41^2} = 507.66 ~ \text{ lb}$

Direction Cosines

$\lambda = \dfrac{{\bf R}}{R} = \dfrac{-90.15{\bf i} - 114{\bf j} + 486.41{\bf k}}{507.66}$

$\lambda = -0.1776{\bf i} - 0.2246{\bf j} + 0.9581{\bf k}$

Calculator Operations: CASIO fx-991ES Plus

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[MODE] → 8:VECTOR → 1:VctA → 1:3
VctA = [ 2 3 4 ]

[AC] → [SHIFT] → [5:VECTOR] → 2:Data → 2:VctB → 1:3
VctB = [ -3 -4 5 ]

[AC] → [SHIFT] → [5:VECTOR] → 2:Data → 3:VctC → 1:3
VctC = [ 0 0 4 ]

Resultant R
R = 100(VctA ÷ Abs(VctA)) + 300(VctB ÷ Abs(VctB)) + 200(VctC ÷ Abs(VctC))

[AC] → 100 → [ ( ] → [SHIFT] → [5:VECTOR] → 3:VctA → [ &divide] → [SHIFT] → [hyp:Abs] → [SHIFT] → [5:VECTOR] → 3:VctA → [ ) ] → [ ) ] → [ + ] → 300 → [ ( ] → [SHIFT] → [5:VECTOR] → 4:VctB → [ &divide] → [SHIFT] → [hyp:Abs] → [SHIFT] → [5:VECTOR] → 4:VctB → [ ) ] → [ ) ] → [ + ] → 200 → [ ( ] → [SHIFT] → [5:VECTOR] → 5:VctC → [ &divide] → [SHIFT] → [hyp:Abs] → [SHIFT] → [5:VECTOR] → 5:VctC → [ ) ] → [ ) ]

R = [ -90.14 -114 486.41 ] answer

For the magnitude of the resultant:

[AC] → [SHIFT] → [hyp:Abs] → [SHIFT] → [5:VECTOR] → 6:VctAns → [ ) ].
R = Abs(VctAns) = 507.66 lb answer

For the direction cosines:

[AC] → [SHIFT] → [5:VECTOR] → 6:VctAns → [ ) ] → ÷ → [SHIFT] → [hyp:Abs] → [SHIFT] → [5:VECTOR] → 6:VctAns → [ ) ].
λ = VctAns ÷ Abs(VctAns)

λ = [ -0.1776 -0.2246 0.9581 ] answer

Problem 603 | Resultant of Concurrent Forces in Space

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