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A 100​-lb load hangs from three cables of equal length that are anchored at the points ​(minus4​,0,0), ​(2​,2 StartRoot 3 EndRoot​,0), and ​(2​,minus2 StartRoot 3 EndRoot​,0). The load is located at ​(0,0,minus4 StartRoot 3 EndRoot​). Find the vectors describing the forces on the cables due to the load.

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fenixbm

Answer:

  • [tex]\vec{F}_{cable_1} =  (-19.245 \ lbf ,0, 33.333 \ lbf)[/tex]
  • [tex]\vec{F}_{cable_2} =  (9.622 \ lbf , 13.608 \ lbf ,33.333 \ lbf)[/tex]
  • [tex]\vec{F}_{cable_3} =  (9.622 \ lbf , -13.608 \ lbf ,33.333 \ lbf)[/tex]

Explanation:

The mass of the load is

[tex]m_{load} = 100 \ lb[/tex]

As the mass hangs, the cables must be tight, so, we can obtain the vector parallel to the cable as:

[tex]\vec{r}_{cable} = \vec{r}_{anchored} - \vec{r}_{load} [/tex]

where [tex]\vec{r}_{load}[/tex] is the position of the load and [tex]\vec{r}_{anchored}[/tex]  is the point where the cable is anchored.

So, for our cables

[tex]\vec{r}_{cable_1} = (-4,0,0) - (0,0,-4\sqrt{3})=(-4,0,4\sqrt{3})[/tex]

[tex]\vec{r}_{cable_2} = (2,2\sqrt{3},0) - (0,0,-4\sqrt{3})=(2,2\sqrt{2},4\sqrt{3})[/tex]

[tex]\vec{r}_{cable_3} = (2,-2\sqrt{3},0) - (0,0,-4\sqrt{3})=(2,-2\sqrt{2},4\sqrt{3})[/tex]

We know that the forces must be in this directions, so we can write

[tex]\vec{F}_i=k_i \vec{r}_{cable_i}[/tex]

We also know, as the system is in equilibrium, the sum of the forces must be zero:

[tex]\vec{F}_{cable_1}+\vec{F}_{cable_2}+\vec{F}_{cable_3}+\vec{W}=0[/tex]

where [tex]\vec{W}[/tex] is the weight,

[tex]\vec{W} = (0,0,-100 \ lbf)[/tex]

So, we get:

[tex]k_1 (-4,0,4\sqrt{3})  + k_2 (2,2\sqrt{2},4\sqrt{3}) + k_3 (2,-2\sqrt{2},4\sqrt{3}) + (0,0,-100 \ lbf) = (0,0,0)[/tex]

This gives us the following equations:

  1. [tex]-4 \ k_1 + 2 \ k_2 + 2 \ k_3  = 0[/tex]
  2. [tex]2\sqrt{2} \ k_2  -2\sqrt{2} \ k_3 = 0[/tex]
  3. [tex]4\sqrt{3} \ k_1   +  4\sqrt{3}  \ k_2 +  4\sqrt{3} \ k_3  -100 \ lbf = 0[/tex]

From equation [2] is clear that [tex]k_2 = k_3[/tex], we can see that

[tex]2\sqrt{2} \ k_2  = 2\sqrt{2} \ k_3[/tex]

[tex]\frac{2\sqrt{2} \ k_2}{2\sqrt{2} }  = \frac{2\sqrt{2} \ k_3}{2\sqrt{2} }[/tex]

[tex]k_2 = k_3[/tex]

Now, putting this in equation [1]

[tex]-4 \ k_1 + 2 \ k_2 + 2 \ k_3  = -4 \ k_1 + 2 \ k_3 + 2 \ k_3 =  -4 \ k_1 + 4 \ k_3  = 0[/tex]

[tex]  4 \ k_1 = 4 \ k_3 [/tex]

[tex]  \ k_1 = \ k_3 [/tex]

Taking this result to the equation [3]

[tex]4\sqrt{3} \ k_1   +  4\sqrt{3}  \ k_2 +  4\sqrt{3} \ k_3  -100 \ lbf = 0[/tex]

[tex]4\sqrt{3}  \ k_3   +  4\sqrt{3}  \ k_3 +  4\sqrt{3} \ k_3  = 100 \ lbf [/tex]

[tex]3 * (4\sqrt{3}  \ k_3)  = 100 \ lbf [/tex]

[tex]  k_3  = \frac{100 \ lbf}{  12 \sqrt{3}} [/tex]

[tex]  k_1 = k_2 = k_3  =  \frac{100 \ lbf}{  12 \sqrt{3}} [/tex]

So, the forces are:

[tex]\vec{F}_{cable_1} = k_1 (-4,0,4\sqrt{3})[/tex]

[tex]\vec{F}_{cable_1} = \frac{100 \ lbf}{  12 \sqrt{3}} (-4,0,4\sqrt{3})[/tex]

[tex]\vec{F}_{cable_1} =  (-\frac{100 \ lbf}{  3 \sqrt{3}},0,\frac{100 \ lbf}{3})[/tex]

[tex]\vec{F}_{cable_1} =  (-19.245 \ lbf ,0, 33.333 \ lbf)[/tex]

[tex]\vec{F}_{cable_2} = k_2 (2,2\sqrt{2},4\sqrt{3})[/tex]

[tex]\vec{F}_{cable_2} = \frac{100 \ lbf}{  12 \sqrt{3}} (2,2\sqrt{2},4\sqrt{3})[/tex]

[tex]\vec{F}_{cable_2} =  (9.622 \ lbf , 13.608 \ lbf ,33.333 \ lbf)[/tex]

[tex]\vec{F}_{cable_3} = k_3 (2,-2\sqrt{2},4\sqrt{3})[/tex]

[tex]\vec{F}_{cable_3} = \frac{100 \ lbf}{  12 \sqrt{3}} (2,-2\sqrt{2},4\sqrt{3})[/tex]

[tex]\vec{F}_{cable_3} =  (9.622 \ lbf , -13.608 \ lbf ,33.333 \ lbf)[/tex]

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