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

To find the vectors describing the forces on the cables due to the load, we can calculate the position vectors from the load to each cable attachment point, find the unit vectors in the direction of each cable, and then multiply the unit vectors by the magnitude of the load to get the forces on the cables.

Step-by-step explanation:

To find the vectors describing the forces on the cables due to the load, we need to calculate the direction and magnitude of each force. The force on a cable can be represented by a vector, where the direction of the vector represents the direction of the force and the magnitude represents the magnitude of the force. To calculate the forces on the cables, we can use the concept of equilibrium, where the sum of the forces in each direction is equal to zero.

First, let's find the position vectors of each cable attachment point in relation to the load. The position vector from the load to the first attachment point is (-4,0,0) - (0,0,-4√3) = (-4,0,4√3). The position vector from the load to the second attachment point is (2, 2√3, 0) - (0,0,-4√3) = (2, 2√3, 4√3). The position vector from the load to the third attachment point is (2, -2√3, 0) - (0,0,-4√3) = (2, -2√3, 4√3).

Next, we can find the unit vectors in the direction of each cable. The unit vector for the first cable is (-4/√148, 0, 4√3/√148) = (-2/√37, 0, √12/√37). The unit vector for the second cable is (2/√37, 2√3/√37, √12/√37). The unit vector for the third cable is (2/√37, -2√3/√37, √12/√37).

Finally, we can find the forces on the cables by multiplying the unit vectors by the magnitude of the load. The force on the first cable is (100 lb) * (-2/√37, 0, √12/√37). The force on the second cable is (100 lb) * (2/√37, 2√3/√37, √12/√37). The force on the third cable is (100 lb) * (2/√37, -2√3/√37, √12/√37).

Answer 2

• 
  `\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)`

Step-by-step 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√(3))=(-4,0,4√(3))`

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

`\vec{r}_(cable_3) = (2,-2√(3),0) - (0,0,-4√(3))=(2,-2√(2),4√(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√(3))  + k_2 (2,2√(2),4√(3)) + k_3 (2,-2√(2),4√(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√(2) \ k_2  -2√(2) \ k_3 = 0`
3. 
   `4√(3) \ k_1   +  4√(3)  \ k_2 +  4√(3) \ k_3  -100 \ lbf = 0`

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

`2√(2) \ k_2  = 2√(2) \ k_3`

`(2√(2) \ k_2)/(2√(2) )  = (2√(2) \ k_3)/(2√(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√(3) \ k_1   +  4√(3)  \ k_2 +  4√(3) \ k_3  -100 \ lbf = 0`

`4√(3)  \ k_3   +  4√(3)  \ k_3 +  4√(3) \ k_3  = 100 \ lbf`

`3 * (4√(3)  \ k_3)  = 100 \ lbf`

`k_3  = (100 \ lbf)/(  12 √(3))`

`k_1 = k_2 = k_3  =  (100 \ lbf)/(  12 √(3))`

So, the forces are:

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

`\vec{F}_(cable_1) = (100 \ lbf)/(  12 √(3)) (-4,0,4√(3))`

`\vec{F}_(cable_1) =  (-(100 \ lbf)/(  3 √(3)),0,(100 \ lbf)/(3))`

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

`\vec{F}_(cable_2) = k_2 (2,2√(2),4√(3))`

`\vec{F}_(cable_2) = (100 \ lbf)/(  12 √(3)) (2,2√(2),4√(3))`

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

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

`\vec{F}_(cable_3) = (100 \ lbf)/(  12 √(3)) (2,-2√(2),4√(3))`

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