A photograph of the NASA Apollo 16 Lunar Module (abbreviated by NASA as the LM is shown on the surface of the Moon. Such spacecraft made six Moon landings during 1,969 - 72. A simplified model for one of the four landing gear assemblies of the LM is shown. If the LM has 13,500 kg mass, and rests on the surface of the Moon where acceleration due to gravity is 1.82 m/s^2, determine the force supported by members AB, AC, and AD. Assume the weight of the LM is uniformly supported by all four landing gear assemblies, and neglect friction between the landing gear and the surface of the Moon. TAB =N TAC = TAD =N A ( 2.6, 2.6, -2.2 ) m B(1.5, 1.5, 0)m C(2,1,-1.2)m D(1,2,-1.2)m

[tex]r_{AB}^{\to} = r _{OA}^{\to} - r_{OB}^{\to} \\ \\ r_{AB}^{\to} = (2.6 \ \hat i + 2.6 \ \hat j - 2.2 \ \hat k ) - ( 1.5 \ \hat i + \ 1. 5 \hat j ) \\ \\ r_{AB}^{\to} = ( 2.6 \ \hat i - 1.5 \ \hat i + 2.6 \ \hat j - 1.5 \ \hat j - 2.2 \ \hat k) \\ \\ r_{AB}^{\to} = (1.1 \ \hat i + 1.1 \ \hat j - 2.2 \ \hat k ) m[/tex]

To calculate the position vector of AC; we have:

[tex]r_{AC}^{\to} = r _{OA}^{\to} - r_{OC}^{\to} \\ \\ r_{AC}^{\to} = (2.6 \ \hat i + 2.6 \ \hat j - 2.2 \ \hat k ) - ( 2\ \hat i + \ \hat j - 1.2 \ \hat k) \\ \\ r_{AC}^{\to} = ( 2.6 \ \hat i - 2\ \hat i + 2.6 \ \hat j - \ \hat j - 2.2 \ \hat k + 1.2 \ \hat k) \\ \\ r_{AC}^{\to} = (0.6 \ \hat i + 1.6 \ \hat j - \ \hat k ) m[/tex]

To calculate the position vector of AD ; we have:

[tex]r_{AD}^{\to} = r _{OA}^{\to} - r_{OD}^{\to} \\ \\ r_{AC}^{\to} = (2.6 \ \hat i + 2.6 \ \hat j - 2.2 \ \hat k ) - ( \hat i + \ 2 \hat j - 1.2 \ \hat k) \\ \\ r_{AD}^{\to} = ( 2.6 \ \hat i - \hat i + 2.6 \ \hat j - 2 \ \hat j - 2.2 \ \hat k + 1.2 \ \hat k) \\ \\ r_{AD}^{\to} = (1.6 \ \hat i + 0.6 \ \hat j - \ \hat k ) m[/tex]

However; let's calculate the force in AB, AC and AD in their respective unit vector form;

To start with unit vector AB by using the following expression; we have:

[tex]F_{AB}^{\to} = F_{AB} \dfrac{ r _{AB}^{\to} }{|r_{AB}^{\to}} \\ \\ \\ F_{AB}^{\to} = F_{AB} \dfrac{(1.1 \ \hat i + 1.1 \ \hat j - 2.2 \ \hat k ) }{\sqrt{ (1.1)^2 + (1.1)^2 + (-2.2 )^2 }} \\ \\ \\ F_{AB}^{\to} = F_{AB} \dfrac{(1.1 \ \hat i + 1.1 \ \hat j - 2.2 \ \hat k ) }{ \sqrt{7.26}} \\ \\ \\ F_{AB}^{\to} = F_{AB} \dfrac{(1.1 \ \hat i + 1.1 \ \hat j - 2.2 \ \hat k ) }{ 2.6944} \\ \\ \\ F_{AB}^{\to} = F_{AB} (0.408 \ \hat i+ 0.408 \ \hat j - 0.8165 \ \hat k ) N\\[/tex]

The force AC in unit vector form is ;

[tex]F_{AC}^{\to} = F_{AC} \dfrac{ r _{AC}^{\to} }{|r_{AC}^{\to}} \\ \\ \\ F_{AC}^{\to} = F_{AC} \dfrac{(0.6 \ \hat i + 1.6 \ \hat j - \ \hat k ) }{\sqrt{ (0.6)^2 + (1.6)^2 + (-1 )^2 }} \\ \\ \\ F_{AC}^{\to} = F_{AC} \dfrac{(0.6 \ \hat i + 1.6 \ \hat j - \ \hat k ) }{ \sqrt{3.92}} \\ \\ \\ F_{AC}^{\to} = F_{AC} \dfrac{(0.6 \ \hat i + 1.6 \ \hat j - \ \hat k ) }{1.9798} \\ \\ \\ F_{AC}^{\to} = F_{AC} (0.303 \ \hat i+ 0.808 \ \hat j - 0.505 \ \hat k ) N\\[/tex]

The force AD in unit vector form is ;

[tex]F_{AD}^{\to} = F_{AD} \dfrac{ r _{AD}^{\to} }{|r_{AD}^{\to}|} \\ \\ \\ F_{AD}^{\to} = F_{AD} \dfrac{(1.6 \ \hat i + 0.6 \ \hat j - \ \hat k ) }{\sqrt{ (1.6)^2 + (0.6)^2 + (-1 )^2 }} \\ \\ \\ F_{AD}^{\to} = F_{AD} \dfrac{(1.6 \ \hat i + 0.6 \ \hat j - \ \hat k ) }{ \sqrt{3.92}} \\ \\ \\ F_{AD}^{\to} = F_{AD} \dfrac{(1.6 \ \hat i + 0.6 \ \hat j - \ \hat k ) }{1.9798} \\ \\ \\ F_{AD}^{\to} = F_{AD} (0.808 \ \hat i+ 0.303 \ \hat j - 0.505 \ \hat k ) N\\[/tex]

Similarly ; the weight of the lunar Module is:

W = mg

where;

mass = 13500 kg

acceleration due to gravity= 1.82 m/s²

W = 13500 × 1.82

W = 24,570 N

Also. we known that the load is shared by four landing gears; Thus, the vertical reaction force exerted by the ground on each landing gear can be expressed as:

[tex]R =\dfrac{W}{4}[/tex]

[tex]R =\dfrac{24,570}{4}[/tex]

R = 6142.5 N

Now; the reaction force at point A in unit vector form is :

[tex]R^{\to} = Rk^{\to} \\ \\ R^{\to} = (6142.5 \ k ^{\to}) \ N[/tex]

Using the force equilibrium at the meeting point of the coordinates at A.

[tex]\sum F^{\to} = 0[/tex]

[tex]F_{AB}^{\to} +F_{AC}^{\to} + F_{AD}^{\to} + R^{\to} =0[/tex]

[tex][F_{AB} (0.408 \ \hat i + 0.408 \ \hat j - 0.8165 \ \hat k ) N + F_{AC} (0.303 \ \hat i + 0.808 \ \hat j - 0.505 \ \hat k ) N + F_{AD} (0.808 \ \hat i + 0.303 \ \hat j - 0.505 \ \hat k) N + (6142.5 \ k^ \to ) ][/tex]

[tex]= [ ( 0.408 F_{AB} +0.303 F_{AC} + 0.808F_{AD}) \hat i + (0.408 F_{AB}+0.808F_{AC}+0.303F_{AD}) \hat j + (-0.8165 F_{AB} -0.505F_{AC} -0.505 F_{AD} +6142.5 ) k ^ \to ] = 0[/tex]

From above; we need to relate and equate each coefficients i.e i ,j, and [tex]k ^ \to[/tex] on both sides ; so, we can re-write that above as;

[tex]0.408 F_{AB} +0.303 F_{AC} + 0.808F_{AD}) =0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ --- (1) \\ \\ 0.408 F_{AB}+0.808F_{AC}+0.303F_{AD}) =0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ --- (2) \\ \\ -0.8165 F_{AB} -0.505F_{AC} -0.505 F_{AD} +6142.5 = 0 --- (3)[/tex]

Making rearrangement and solving by elimination method;

[tex]\mathbf{F_{AB} = 13785.06 N }[/tex]

[tex]\mathbf{F_{AC} = -5062.38 N }[/tex]

[tex]\mathbf{F_{AD} = -5062.38 N }[/tex]

oeerivona oeerivona · Answer 2 · 2022-02-25T20:55:45+01:00

The force vector of each member, depends on the magnitude of the

force and the unit vector of the member.

Responses:

The force supported by the members are;

Force supported by AB is; 13,799.95 N

Force supported by AC is; -5070.2 N

Force supported by AD is -5070.2 N

How can the unit vector of each member give their force?

Resolving the given members into unit vectors gives;

[tex]\hat u_{AB} = \mathbf{\dfrac{(2.6 - 1.5) \cdot \hat i + (2.6 - 1.5)\cdot \hat j + (-2.2)\cdot \hat k }{\sqrt{(2.6- 1.5)^2 + (2.6 - 1.5)^2 + (-2.2)^2}}}[/tex][tex]\dfrac{(2.6 - 1.5) \cdot \hat i + (2.6 - 1.5)\cdot \hat j + (-2.2)\cdot \hat k }{\sqrt{(2.6- 1.5)^2 + (2.6 - 1.5)^2 + (-2.2)^2}}= 0.40825 \cdot \hat i + 0.40825\cdot \hat j - 0.81625\cdot \hat k[/tex]

[tex]\hat u_{AB} = \mathbf{0.40825 \cdot \hat i + 0.40825\cdot \hat j - 0.81625\cdot \hat k}[/tex]

Similarly, we have;

[tex]\hat u_{AC} =\mathbf{ \dfrac{(2.6 - 2) \cdot \hat i + (2.6 - 1)\cdot \hat j + (-2.2+1.2)\cdot \hat k }{\sqrt{(2.6- 2)^2 + (2.6 - 1)^2 + (-2.2+1.2)^2}}}[/tex]

[tex]\dfrac{(2.6 - 2) \cdot \hat i + (2.6 - 1)\cdot \hat j + (-2.2+1.2)\cdot \hat k }{\sqrt{(2.6- 2)^2 + (2.6 - 1)^2 + (-2.2+1.2)^2}} =\dfrac{0.6\cdot \hat i +1.6\cdot \hat j -1\cdot \hat k }{\sqrt{0.6^2 + 1.6^2 + (-1.)^2}}[/tex]

[tex]\dfrac{0.6\cdot \hat i +1.6\cdot \hat j -1\cdot \hat k }{\sqrt{0.6^2 + 1.6^2 + (-1.)^2}}= 0.303046\cdot \hat i + 0.80812\cdot \hat j - 0.50508\cdot \hat k[/tex]

[tex]\hat u_{AC} =\mathbf{0.303046\cdot \hat i + 0.80812\cdot \hat j - 0.50508\cdot \hat k}[/tex]

[tex]\hat u_{AD} =\mathbf{ \dfrac{(2.6 - 1) \cdot \hat i + (2.6 -2)\cdot \hat j + (-2.2 + 1.2)\cdot \hat k }{\sqrt{(2.6-1)^2 + (2.6 -2))^2 + (-2.2 + 1.2)^2}}}[/tex]

[tex]\hat u_{AD} =\mathbf{0.80812\cdot \hat i+ 0.303046\cdot \hat j - 0.50508\cdot \hat k}[/tex]

The forces are therefore;

[tex]\vec F_{AB} =\mathbf{ F_{AB} \cdot \left ( 0.40825 \cdot \hat i + 0.40825\cdot \hat j - 0.81625\cdot \hat k \right)}[/tex]

[tex]\vec F_{AC} =\mathbf{ F_{AC} \cdot \left (0.303046\cdot \hat i + 0.80812\cdot \hat j - 0.50508\cdot \hat k\right)}[/tex]

[tex]\vec F_{AD} = \mathbf{F_{AD} \cdot \left (0.80812\cdot \hat i+ 0.303046\cdot \hat j - 0.50508\cdot \hat k\right)}[/tex]

[tex]Weight \ on \ the \ assembly = \dfrac{13,500 \, kg \times 1.82 \, m/s^2}{4} = 6,142.5 \, \hat k N[/tex]

Which gives;

[tex]\mathbf{0.40825 \cdot \hat i \cdot F_{AB}}[/tex] + [tex]0.303046\cdot \hat i \cdot F_{AC}[/tex] + [tex]0.80812\cdot \hat i \cdot F_{AD}[/tex] = 0

[tex]0.40825 \cdot \hat j \cdot F_{AB}[/tex] + [tex]0.80812\cdot \hat j \cdot F_{AC}[/tex] + [tex]0.303046 \cdot \hat j \cdot F_{AD}\left[/tex] = 0

[tex]-0.81625\cdot \hat k \cdot F_{AB}[/tex] - [tex]0.50508\cdot \hat k \cdot F_{AC}[/tex] - [tex]0.50508\cdot \hat k \cdot F_{AD}[/tex] + [tex]\mathbf{6,142.5 \, \hat k}[/tex] = 0

Which gives;

[tex]-0.81625\cdot \hat k \cdot F_{AB}[/tex] - [tex]0.50508\cdot \hat k \cdot F_{AC}[/tex] - [tex]0.50508\cdot \hat k \cdot F_{AD}[/tex] = [tex]-6,142.5 \, \hat k[/tex]

Solving gives;

[tex]F_{AB}[/tex] = 13799.95 N

[tex]F_{AC}[/tex] = -5070.2 N

[tex]F_{AD}[/tex] = -5070.2 N

Learn more about unit vectors here:

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