Question

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.

Answer 1

Answer:

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

Explanation:

The mass of the load is

$m_{load} = 100 \ lb$

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

$\vec{r}_{cable} = \vec{r}_{anchored} - \vec{r}_{load}$

where $\vec{r}_{load}$ is the position of the load and $\vec{r}_{anchored}$  is the point where the cable is anchored.

So, for our cables

$\vec{r}_{cable_1} = (-4,0,0) - (0,0,-4\sqrt{3})=(-4,0,4\sqrt{3})$

$\vec{r}_{cable_2} = (2,2\sqrt{3},0) - (0,0,-4\sqrt{3})=(2,2\sqrt{2},4\sqrt{3})$

$\vec{r}_{cable_3} = (2,-2\sqrt{3},0) - (0,0,-4\sqrt{3})=(2,-2\sqrt{2},4\sqrt{3})$

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

$\vec{F}_i=k_i \vec{r}_{cable_i}$

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

$\vec{F}_{cable_1}+\vec{F}_{cable_2}+\vec{F}_{cable_3}+\vec{W}=0$

where $\vec{W}$ is the weight,

$\vec{W} = (0,0,-100 \ lbf)$

So, we get:

$k_1 (-4,0,4\sqrt{3}) + k_2 (2,2\sqrt{2},4\sqrt{3}) + k_3 (2,-2\sqrt{2},4\sqrt{3}) + (0,0,-100 \ lbf) = (0,0,0)$

This gives us the following equations:

1. $-4 \ k_1 + 2 \ k_2 + 2 \ k_3 = 0$
2. $2\sqrt{2} \ k_2 -2\sqrt{2} \ k_3 = 0$
3. $4\sqrt{3} \ k_1 + 4\sqrt{3} \ k_2 + 4\sqrt{3} \ k_3 -100 \ lbf = 0$

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

$2\sqrt{2} \ k_2 = 2\sqrt{2} \ k_3$

$\frac{2\sqrt{2} \ k_2}{2\sqrt{2} } = \frac{2\sqrt{2} \ k_3}{2\sqrt{2} }$

$k_2 = k_3$

Now, putting this in equation [1]

$-4 \ k_1 + 2 \ k_2 + 2 \ k_3 = -4 \ k_1 + 2 \ k_3 + 2 \ k_3 = -4 \ k_1 + 4 \ k_3 = 0$

$4 \ k_1 = 4 \ k_3$

$\ k_1 = \ k_3$

Taking this result to the equation [3]

$4\sqrt{3} \ k_1 + 4\sqrt{3} \ k_2 + 4\sqrt{3} \ k_3 -100 \ lbf = 0$

$4\sqrt{3} \ k_3 + 4\sqrt{3} \ k_3 + 4\sqrt{3} \ k_3 = 100 \ lbf$

$3 * (4\sqrt{3} \ k_3) = 100 \ lbf$

$k_3 = \frac{100 \ lbf}{ 12 \sqrt{3}}$

$k_1 = k_2 = k_3 = \frac{100 \ lbf}{ 12 \sqrt{3}}$

So, the forces are:

$\vec{F}_{cable_1} = k_1 (-4,0,4\sqrt{3})$

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

$\vec{F}_{cable_1} = (-\frac{100 \ lbf}{ 3 \sqrt{3}},0,\frac{100 \ lbf}{3})$

$\vec{F}_{cable_1} = (-19.245 \ lbf ,0, 33.333 \ lbf)$

$\vec{F}_{cable_2} = k_2 (2,2\sqrt{2},4\sqrt{3})$

$\vec{F}_{cable_2} = \frac{100 \ lbf}{ 12 \sqrt{3}} (2,2\sqrt{2},4\sqrt{3})$

$\vec{F}_{cable_2} = (9.622 \ lbf , 13.608 \ lbf ,33.333 \ lbf)$

$\vec{F}_{cable_3} = k_3 (2,-2\sqrt{2},4\sqrt{3})$

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

$\vec{F}_{cable_3} = (9.622 \ lbf , -13.608 \ lbf ,33.333 \ lbf)$