Question

Determine the x, y, z components of reaction at the ball-and-socket joint A.

Determine the x, y, z components of the moments where the wing is fixed to the fuselage A.
The sign has a mass of 90kg with center of mass at G.
(Figure 1)

Answer 1

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Hello! In this question, I will answer the first part of the question, in which we will determine the x, y, and z components for the reaction at joint A in the ball-and-socket.
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Step-by-step explanation:

To better understand what the question entails, we will start off by drawing the free-body diagram to understand the direction of our components. The image is attached below. This will help us solve for our components.

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Solve:

To start, we will first solve for the weight of the sign, we will use the formula:

`W=mg`

Whereas:

• m = mass (90)
• g = gravity (9.81)

Plug in our values into the formula and solve:

`W = 90\cdot9.81=882.9 N`

Since we now know our weight value, we can solve for our force in the cable BC, expressed as a vector. We will use the formula:

`\bold{T}_(BC)=T_(BC)(\frac{\bold{r}_(BC)}{r_(BC)})`

Where:

• 
  `\bold{r}_{BC` is coordinate B subtracted from the coordinate C
• 
  `r_(BC)` is the magnitude of BC

Plug in our values into the formula and solve:

`\bold{T}_(BC)=T_(BC)\bigg(\cfrac{\text{[i - 2j+ 2k] ft}}{√((1)^2+(-2)^2+(2)^2)}}\bigg)\\\\\bold{T}_(BC)=T_(BC)\bigg(\cfrac{\text{[i - 2j+ 2k] ft}}{3}}}\bigg)\\\\\text{Simplify}\\\\\bold{T}_(BC)=(1)/(3)T_(BC){\text{i}}-(2)/(3)T_(BC)j+(2)/(3) T_(BC)k`

Now, let us solve for our force in the cable BD, also expressed as a vector. Use the formula:

`\bold{T}_(BD)=T_(BD)(\frac{\bold{r}_(BD)}{r_(BD)})`

Use the same steps from solving for our vector force of cable BC. Plug in the values and solve:

`\bold{T}_(BD)=T_(BD)\bigg(\cfrac{\text{[-2i - 2j+ k] ft}}{√((-2)^2+(-2)^2+(1)^2)}}\bigg)\\\\\bold{T}_(BD)=T_(BD)\bigg(\cfrac{\text{[-2i - 2j+ k] ft}}{3}}}\bigg)\\\\\text{Simplify}\\\\\bold{T}_(BD)=-(2)/(3)T_(BD){\text{i}}-(2)/(3)T_(BD)j+(1)/(3) T_(BD)k`

We now need to find the moment at A at equilibrium (0). This is known as:

`\sum M_(a)=0 \Rightarrow\bold{r}_(B)*(\bold{T}_(BC)+\bold{T}_(BD)+\bold{W})=0`

Whereas:

• 
  `\bold{T}_(BC)=\big((1)/(3)T_(BC){\text{i}}-(2)/(3)T_(BC)j+(2)/(3) T_(BC)k\big)`
• 
  `\bold{T}_(BD)=\big(-(2)/(3)T_(BD){\text{i}}-(2)/(3)T_(BD)j+(1)/(3) T_(BD)k\big)`
• 
  `\bold{W}=-882.9\text{k}`
• 
  `\bold{r}_(B)=2\text{j}`

Plug in values into the equation:

`2\text{j}*\big[\big((1)/(3)T_(BC){\text{i}}-(2)/(3)T_(BC)j+(2)/(3) T_(BC)k\big)+\big(-(2)/(3)T_(BD){\text{i}}-(2)/(3)T_(BD)j+(1)/(3) T_(BD)k\big)\big]\\+\text{j}*-882.9\text{k}=0`

Find the moment about the z-axis to zero. This is known as:

`\sum M_(z)=0\\\\-(2)/(3)T_(BC)+(4)/(3)T_(BD)=0\\\\\text{Simplify}\\\\T_(BC)=2T_(BD)`

Now, find the moment about the x-axis to zero. This is known as:

`\sum M_(x)=0\\\\(4)/(3)T_(BC)+(2)/(3)T_(BD)-882.9=0\\\\\text{We know }T_(BC)=2T_(BD)\text{, plug in }2T_(BD)\text{ in }T_(BC)\\\\(4)/(3)*2T_(BD)+(2)/(3)T_(BD)-882.9=0\text{ Solve for }T_(BD)\\\\(10)/(3)T_(BD)=882.9\\\\T_(BD)=264.87\text{ N}`

Now, let us calculate the tension in wire BC.

`T_(BC)=2T_(BD) = 2(264.87) = 529.74\text{ N}`

Let us calculate the reaction at A by solving for the equilibrium force equation. The formula is:

`\sum F=0\\\\\bold{F}_A+\bold{T}_(BC)+\bold{T}_(BD)+\bold{W}=0`

Plug in our known information into the equation and simplify.

`\big[(A_xi+A_yj+A_zk)+((1)/(3)T_(BC){\text{i}}-(2)/(3)T_(BC)j+(2)/(3) T_(BC)k)+(-(2)/(3)T_(BD){\text{i}}-(2)/(3)T_(BD)j\\+(1)/(3) T_(BD)k)-981\text{k}\big]=0\\\\\text{Separate by component}\\\\\big[(A_x+(1)/(3)T_(BC)-(2)/(3)T_(BD))\text{\bold{i}}+(A_y-(2)/(3)T_(BC)-(2)/(3)T_(BD))\text{\bold{j}}+(A_z+(2)/(3)T_(BC)+(1)/(3)T_(BD)\\-981)\text{\bold{k}}\big]=0\\\\\text{Plug in known values}\\`

`\big[(A_x+(1)/(3)*529.74-(2)/(3)*264.87)\text{\bold{i}}+(A_y-(2)/(3)*529.74-(2)/(3)*264.87)\text{\bold{j}}+(A_z+(2)/(3)*529.74+(1)/(3)*264.87-882.9)\text{\bold{k}}\big]=0`

Now, we can calculate our reactions.

Calculate the reaction at A in the x-direction.

`\sum F_x=0\\\\A_x+(1)/(3)*529.74-(2)/(3)*264.87=0\\\\\boxed{A_x=0\text{ N}}`

Calculate the reaction at A in the y-direction.

`\sum F_y=0\\\\A_y-(2)/(3)*529.74-(2)/(3)*264.87=0\\\\A_y-529.74=0\\\\\boxed{A_y=529.74\text{ N}}`

Calculate the reaction at A in the z-direction.

`\sum F_z=0\\\\A_z+(2)/(3)*529.74+(1)/(3)*264.87-981=0\\\\A_z-441.45=0\\\\\boxed{A_z=441.45\text{ N}}`

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Answer:

`\boxed{\text{A=} < \text{0, 529.74, 441.45} > }`

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