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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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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:
[tex]W=mg[/tex]
Whereas:
- m = mass (90)
- g = gravity (9.81)
Plug in our values into the formula and solve:
[tex]W = 90\cdot9.81=882.9 N[/tex]
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:
[tex]\bold{T}_{BC}=T_{BC}(\frac{\bold{r}_{BC}}{r_{BC}})[/tex]
Where:
- [tex]\bold{r}_{BC[/tex] is coordinate B subtracted from the coordinate C
- [tex]r_{BC}[/tex] is the magnitude of BC
Plug in our values into the formula and solve:
[tex]\bold{T}_{BC}=T_{BC}\bigg(\cfrac{\text{[i - 2j+ 2k] ft}}{\sqrt{(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}=\frac{1}{3}T_{BC}{\text{i}}-\frac{2}{3}T_{BC}j+\frac{2}{3} T_{BC}k[/tex]
Now, let us solve for our force in the cable BD, also expressed as a vector. Use the formula:
[tex]\bold{T}_{BD}=T_{BD}(\frac{\bold{r}_{BD}}{r_{BD}})[/tex]
Use the same steps from solving for our vector force of cable BC. Plug in the values and solve:
[tex]\bold{T}_{BD}=T_{BD}\bigg(\cfrac{\text{[-2i - 2j+ k] ft}}{\sqrt{(-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}=-\frac{2}{3}T_{BD}{\text{i}}-\frac{2}{3}T_{BD}j+\frac{1}{3} T_{BD}k[/tex]
We now need to find the moment at A at equilibrium (0). This is known as:
[tex]\sum M_{a}=0 \Rightarrow\bold{r}_{B}\times(\bold{T}_{BC}+\bold{T}_{BD}+\bold{W})=0[/tex]
Whereas:
- [tex]\bold{T}_{BC}=\big(\frac{1}{3}T_{BC}{\text{i}}-\frac{2}{3}T_{BC}j+\frac{2}{3} T_{BC}k\big)[/tex]
- [tex]\bold{T}_{BD}=\big(-\frac{2}{3}T_{BD}{\text{i}}-\frac{2}{3}T_{BD}j+\frac{1}{3} T_{BD}k\big)[/tex]
- [tex]\bold{W}=-882.9\text{k}[/tex]
- [tex]\bold{r}_{B}=2\text{j}[/tex]
Plug in values into the equation:
[tex]2\text{j}\times\big[\big(\frac{1}{3}T_{BC}{\text{i}}-\frac{2}{3}T_{BC}j+\frac{2}{3} T_{BC}k\big)+\big(-\frac{2}{3}T_{BD}{\text{i}}-\frac{2}{3}T_{BD}j+\frac{1}{3} T_{BD}k\big)\big]\\+\text{j}\times-882.9\text{k}=0[/tex]
Find the moment about the z-axis to zero. This is known as:
[tex]\sum M_{z}=0\\\\-\frac{2}{3}T_{BC}+\frac{4}{3}T_{BD}=0\\\\\text{Simplify}\\\\T_{BC}=2T_{BD}[/tex]
Now, find the moment about the x-axis to zero. This is known as:
[tex]\sum M_{x}=0\\\\\frac{4}{3}T_{BC}+\frac{2}{3}T_{BD}-882.9=0\\\\\text{We know }T_{BC}=2T_{BD}\text{, plug in }2T_{BD}\text{ in }T_{BC}\\\\\frac{4}{3}\times2T_{BD}+\frac{2}{3}T_{BD}-882.9=0\text{ Solve for }T_{BD}\\\\\frac{10}{3}T_{BD}=882.9\\\\T_{BD}=264.87\text{ N}[/tex]
Now, let us calculate the tension in wire BC.
[tex]T_{BC}=2T_{BD} = 2(264.87) = 529.74\text{ N}[/tex]
Let us calculate the reaction at A by solving for the equilibrium force equation. The formula is:
[tex]\sum F=0\\\\\bold{F}_A+\bold{T}_{BC}+\bold{T}_{BD}+\bold{W}=0[/tex]
Plug in our known information into the equation and simplify.
[tex]\big[(A_xi+A_yj+A_zk)+(\frac{1}{3}T_{BC}{\text{i}}-\frac{2}{3}T_{BC}j+\frac{2}{3} T_{BC}k)+(-\frac{2}{3}T_{BD}{\text{i}}-\frac{2}{3}T_{BD}j\\+\frac{1}{3} T_{BD}k)-981\text{k}\big]=0\\\\\text{Separate by component}\\\\\big[(A_x+\frac{1}{3}T_{BC}-\frac{2}{3}T_{BD})\text{\bold{i}}+(A_y-\frac{2}{3}T_{BC}-\frac{2}{3}T_{BD})\text{\bold{j}}+(A_z+\frac{2}{3}T_{BC}+\frac{1}{3}T_{BD}\\-981)\text{\bold{k}}\big]=0\\\\\text{Plug in known values}\\[/tex]
[tex]\big[(A_x+\frac{1}{3}\times529.74-\frac{2}{3}\times264.87)\text{\bold{i}}+(A_y-\frac{2}{3}\times529.74-\frac{2}{3}\times264.87)\text{\bold{j}}+(A_z+\frac{2}{3}\times529.74+\frac{1}{3}\times264.87-882.9)\text{\bold{k}}\big]=0[/tex]
Now, we can calculate our reactions.
Calculate the reaction at A in the x-direction.
[tex]\sum F_x=0\\\\A_x+\frac{1}{3}\times529.74-\frac{2}{3}\times264.87=0\\\\\boxed{A_x=0\text{ N}}[/tex]
Calculate the reaction at A in the y-direction.
[tex]\sum F_y=0\\\\A_y-\frac{2}{3}\times529.74-\frac{2}{3}\times264.87=0\\\\A_y-529.74=0\\\\\boxed{A_y=529.74\text{ N}}[/tex]
Calculate the reaction at A in the z-direction.
[tex]\sum F_z=0\\\\A_z+\frac{2}{3}\times529.74+\frac{1}{3}\times264.87-981=0\\\\A_z-441.45=0\\\\\boxed{A_z=441.45\text{ N}}[/tex]
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Answer:
[tex]\boxed{\text{A=} < \text{0, 529.74, 441.45} > }[/tex]
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