Problem 604 | Resultant of Concurrent Forces in Space | Engineering Mechanics Review at MATHalino

Problem 604
Determine the magnitude of the resultant, its pointing, and its direction cosines for the following system of non-coplanar concurrent forces. 200 lb (4, 5, –3); 400 lb (–6, 4, –5); 300 lb, (4, –2, –3).

Solution 604

Click here to expand or collapse this section

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
200	4	5	-3	7.0711	113.1371	141.4214	-84.8528
400	–6	4	-5	8.7750	-273.5054	182.3370	-227.9212
300	4	-2	-3	5.3852	222.8344	-111.4172	-167.1258
SUM (R_x, R_y, and R_z)					62.4661	212.3412	-479.8998
RESULTANT					R = 528.4833

Direction cosines of the resultant

$\cos \theta_x = \dfrac{R_x}{R} = 0.1182$

$\cos \theta_y = \dfrac{R_y}{R} = 0.4018$

$\cos \theta_z = \dfrac{R_z}{R} = -0.9081$

Another Solution (Using Vectors)

Click here to expand or collapse this section

Direction vectors

${\bf r}_1 = 4{\bf i} + 5{\bf j} - 3{\bf k}$

${\bf r}_2 = -6{\bf i} + 4{\bf j} - 5{\bf k}$

${\bf r}_3 = 4{\bf i} - 2{\bf j} - 3{\bf k}$

Unit vectors

${\bf \lambda}_1 = \dfrac{ 4{\bf i} + 5{\bf j} - 3{\bf k}}{\sqrt{4^2 + 5^2 + 3^2}} = 0.5657{\bf i} + 0.7071{\bf j} - 0.4243{\bf k}$

${\bf \lambda}_2 = \dfrac{-6{\bf i} + 4{\bf j} - 5{\bf k}}{\sqrt{6^2 + 4^2 + 5^2}} = -0.6838{\bf i} + 0.4558{\bf j} - 0.5698{\bf k}$

${\bf \lambda}_3 = \dfrac{4{\bf i} - 2{\bf j} - 3{\bf k}}{\sqrt{4^2 + 2^2 + 3^2}} = 0.7428{\bf i} - 0.3714{\bf j} - 0.5571{\bf k}$

Forces in rectangular form

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

${\bf F}_1 = 200(0.5657{\bf i} + 0.7071{\bf j} - 0.4243{\bf k}) = 113.14{\bf i} + 141.42{\bf j} - 84.86{\bf k} ~ \text{lb}$

${\bf F}_2 = 400(-0.6838{\bf i} + 0.4558{\bf j} - 0.5698{\bf k}) = -273.52{\bf i} + 182.32{\bf j} - 227.92{\bf k} ~ \text{lb}$

${\bf F}_3 = 300(0.7428{\bf i} - 0.3714{\bf j} - 0.5571{\bf k}) = 222.84{\bf i} - 111.42{\bf j} - 167.13{\bf k} ~ \text{lb}$

Resultant

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

Sum up the coefficients of i, j and k
${\bf R} = 62.46{\bf i} + 212.32{\bf j} - 479.91{\bf k} ~ \text{lb}$ answer

Magnitude of resultant

$R = \sqrt{62.46^2 + 212.32^2 + 479.91^2} = 528.48 ~ \text{ lb}$

Direction Cosines

$\lambda = \dfrac{{\bf R}}{R} = \dfrac{62.46{\bf i} + 212.32{\bf j} - 479.91{\bf k}}{528.48}$

$\lambda = 0.1182{\bf i} + 0.4018{\bf j} - 0.9081{\bf k}$

Calculator Operations: CASIO fx-991ES Plus

Click here to expand or collapse this section

[MODE] → 8:VECTOR → 1:VctA → 1:3
VctA = [ 4 5 -3 ]

[AC] → [SHIFT] → [5:VECTOR] → 2:Data → 2:VctB → 1:3
VctB = [ -6 4 -5 ]

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

Resultant R
R = 200(VctA ÷ Abs(VctA)) + 400(VctB ÷ Abs(VctB)) + 300(VctC ÷ Abs(VctC))

[AC] → 200 → [ ( ] → [SHIFT] → [5:VECTOR] → 3:VctA → [ &divide] → [SHIFT] → [hyp:Abs] → [SHIFT] → [5:VECTOR] → 3:VctA → [ ) ] → [ ) ] → [ + ] → 400 → [ ( ] → [SHIFT] → [5:VECTOR] → 4:VctB → [ &divide] → [SHIFT] → [hyp:Abs] → [SHIFT] → [5:VECTOR] → 4:VctB → [ ) ] → [ ) ] → [ + ] → 300 → [ ( ] → [SHIFT] → [5:VECTOR] → 5:VctC → [ &divide] → [SHIFT] → [hyp:Abs] → [SHIFT] → [5:VECTOR] → 5:VctC → [ ) ] → [ ) ]

R = [ 62.46 212.32 -479.91 ] answer

For the magnitude of the resultant:

[AC] → [SHIFT] → [hyp:Abs] → [SHIFT] → [5:VECTOR] → 6:VctAns → [ ) ].
R = Abs(VctAns) = 528.48 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 = [ 0.1182 0.4018 -0.9081 ] answer

Problem 604 | Resultant of Concurrent Forces in Space

Navigation

Recent comments