Question

2.102 three cables are used to tether a balloon as shown. knowing that the balloon exerts an 800-n vertical force at a, determine the tension in each cable. fig. p2.101 and p2.102 (diagram shows balloon at a, points b, c, d, o, with lengths: 2.40 m, 4.20 m, 3.30 m, 5.60 m, and axes x, y, z)

Explanation:

Step1: Determine the length of each cable

First, we need to find the length of each cable (AB, AC, AD) using the distance formula in 3D. Let's assume point O is at the origin \((0,0,0)\) and point A is at \((0,5.60,0)\) (since the vertical distance from O to A is 5.60 m, and we can infer the x and z coordinates for each point:

- For cable AB: Point B is at \((-4.20, 0, 4.20)\)? Wait, no, looking at the diagram, let's get the coordinates right. Wait, the horizontal distances: For B, maybe x and z? Wait, the diagram has for B: 420 m (maybe x or z) and for C: 2.40 m (z) and 4.20 m (x)? Wait, maybe better to get the coordinates of each point:

Let’s define the coordinates:

- Point O: \((0, 0, 0)\)
- Point A: \((0, 5.60, 0)\) (since the vertical distance from O to A is 5.60 m along the y-axis)
- Point B: Let's see, the horizontal distances: From the diagram, for B, the x-distance from O is -4.20 m (assuming left), z-distance is 4.20 m? Wait, no, maybe:

Wait, the diagram shows for B: 420 m (maybe x) and 420 m (z)? Wait, no, the labels:

- For cable AB: The horizontal component (in x-z plane) distance from O to B: let's calculate the length of OB. From the diagram, OB seems to be a diagonal? Wait, no, maybe the coordinates:

Wait, let's re-express:

- Cable AB: Point B is at \((-4.20, 0, 4.20)\)? Wait, no, the horizontal distance (in x-z plane) from O to B: let's compute the length of OB. If we look at the x and z coordinates: for B, x is -4.20 m, z is 4.20 m? Then OB length is \(\sqrt{(-4.20)^2 + 4.20^2} = \sqrt{2 \times 4.20^2} = 4.20\sqrt{2}\) m. Then the length of AB is \(\sqrt{(4.20\sqrt{2})^2 + 5.60^2}\). Wait, no, AB is from A(0,5.60,0) to B(x,y,z) where B is at (x,0,z) (since it's on the horizontal plane, y=0). So the length of AB is \(\sqrt{(x - 0)^2 + (0 - 5.60)^2 + (z - 0)^2} = \sqrt{x^2 + z^2 + 5.60^2}\).

Wait, let's get the correct coordinates:

- Point B: From the diagram, the x-distance from O is -4.20 m (left), z-distance is 4.20 m (front)? Wait, no, the labels:

Wait, the problem says:

- For cable AB: The horizontal (x-z) distance from O to B: let's see the diagram, the length from O to B in x-z plane: maybe 4.20 m in x and 4.20 m in z? Wait, no, the diagram has "420 m" for B, "330 m" for D, "2.40 m" and "4.20 m" for C.

Wait, let's parse the diagram:

- Cable AB: The horizontal (x-z) component: let's say point B is at ( -4.20, 0, 4.20 ) m (so x=-4.20, z=4.20, y=0). Then the distance from O to B is \(\sqrt{(-4.20)^2 + 4.20^2} = \sqrt{2 \times 4.20^2} = 4.20\sqrt{2}\) m. Then the length of AB is \(\sqrt{(4.20\sqrt{2})^2 + 5.60^2}\). Let's calculate that:

\( (4.20\sqrt{2})^2 = 4.20^2 \times 2 = 17.64 \times 2 = 35.28 \)

\( 5.60^2 = 31.36 \)

So AB length: \(\sqrt{35.28 + 31.36} = \sqrt{66.64} \approx 8.163\) m? Wait, no, that can't be. Wait, maybe the vertical distance is 5.60 m (along y-axis), and the horizontal distance (x-z) for AB is 7.0 m? Wait, 4.20 and 5.60? Wait, 4.2-5.6-7.0 is a 3-4-5 triangle scaled by 1.4: 4.2=3×1.4, 5.6=4×1.4, so 7.0=5×1.4. Ah! That makes sense. So maybe the horizontal distance (x-z) for AB is 7.0 m? Wait, 4.20 and 5.60? Wait, 4.20 is 3×1.4, 5.60 is 4×1.4, so the hypotenuse (horizontal) would be 5×1.4=7.0 m. Then the length of AB is \(\sqrt{7.0^2 + 5.60^2}\)? No, wait, AB is from A(0,5.60,0) to B(x,0,z), so the vertical component is 5.60 m (along y-axis), and the horizontal component (x-z) is the distance from O to B, which is \(\sqrt{x^2 + z^2}\). If the horizontal component is 7.0 m (since 4.2 and 5.6 are 3:4, so hypotenuse 5×1.4=7), then AB length is \(\sqrt{7.0^2 + 5.60^2}\)? Wait, no, 5.60 is the vertical distance (y-axis), so the length of AB is \(\sqrt{(x - 0)^2 + (0 - 5.60)^2 + (z - 0)^2} = \sqrt{x^2 + z^2 + 5.60^2}\). If x^2 + z^2 = 7.0^2 = 49, then AB length is \(\sqrt{49 + 31.36} = \sqrt{80.36} \approx 8.964\) m? Wait, no, I think I messed up the coordinates.

Wait, let's start over. Let's define the coordinates properly:

- Let O be at (0, 0, 0).
- Point A is at (0, 5.60, 0) (since the vertical distance from O to A is 5.60 m along the y-axis).
- Point B: Let's look at the diagram, the horizontal (x-z) distance from O to B: the x-coordinate is -4.20 m (left), z-coordinate is 4.20 m (front)? Wait, no, the diagram shows "420 m" for B, "330 m" for D, "2.40 m" and "4.20 m" for C.

Wait, maybe:

- Cable AB: The horizontal (x-z) component: x = -4.20 m, z = 4.20 m (so OB vector is (-4.20, 0, 4.20)). Then the length of OB is \(\sqrt{(-4.20)^2 + 4.20^2} = \sqrt{2 \times 4.20^2} = 4.20\sqrt{2} \approx 5.94\) m. Then the length of AB is \(\sqrt{(4.20\sqrt{2})^2 + 5.60^2} = \sqrt{35.28 + 31.36} = \sqrt{66.64} \approx 8.16\) m.

- Cable AC: Point C is at (4.20, 0, -2.40) (x=4.20, z=-2.40). Then OB (OC) length is \(\sqrt{4.20^2 + (-2.40)^2} = \sqrt{17.64 + 5.76} = \sqrt{23.4} \approx 4.837\) m. Then AC length is \(\sqrt{4.837^2 + 5.60^2} \approx \sqrt{23.4 + 31.36} = \sqrt{54.76} \approx 7.4\) m? Wait, no, 4.20^2 + 2.40^2 = 17.64 + 5.76 = 23.4, square root is ~4.837, then 4.837^2 + 5.60^2 = 23.4 + 31.36 = 54.76, square root is 7.4 m (since 7.4^2 = 54.76). Ah, that's a nice number.

- Cable AD: Point D is at (3.30, 0, -4.40)? Wait, no, the diagram shows "330 m" and "5.60 m"? Wait, no, the horizontal (x-z) for D: x=3.30, z=-4.40? Wait, 3.3-4.4-5.5 is a 3-4-5 triangle scaled by 1.1: 3.3=3×1.1, 4.4=4×1.1, so 5.5=5×1.1. Then the length of OD is 5.5 m, and AD length is \(\sqrt{5.5^2 + 5.60^2}\)? Wait, no, 5.5^2 = 30.25, 5.60^2=31.36, so AD length is \(\sqrt{30.25 + 31.36} = \sqrt{61.61} \approx 7.85\) m? No, that doesn't fit. Wait, maybe the vertical distance is 5.60 m, and the horizontal distances (x-z) for each cable are:

- AB: horizontal distance (OD) = 7.0 m (since 4.2-5.6-7.0 is 3-4-5×1.4), so AB length = \(\sqrt{7.0^2 + 5.60^2}\)? Wait, 7.0^2=49, 5.60^2=31.36, sum=80.36, sqrt=8.964≈9.0 m? No, 7-5.6-8.96 is 5-4-6.4, which is a similar triangle.

Wait, maybe the key is that the vertical force at A is 800 N upward (wait, no, the balloon exerts a vertical force at A, probably downward? Wait, the balloon is tethered, so the force at A is downward (weight), and the cables pull upward. So we have three cables, each with tension T_AB, T_AC, T_AD, and the vertical component of each tension must sum to 800 N (since the system is in equilibrium, vertical forces balance: sum of vertical components of tensions = 800 N downward? Wait, no: if the balloon exerts a 800 N vertical force downward at A, then the sum of the vertical components of the cable tensions must be 800 N upward.

To find the vertical component of each tension, we need the unit vector in the direction of each cable, then take the y-component (since the vertical direction is y).

First, let's find the coordinates of each point:

- Point A: (0, 5.60, 0)
- Point B: Let's assume from the diagram, the horizontal (x-z) coordinates of B are (-4.20, 0, 4.20) (so x=-4.20, z=4.20, y=0)
- Point C: (4.20, 0, -2.40) (x=4.20, z=-2.40, y=0)
- Point D: (3.30, 0, -4.40) (x=3.30, z=-4.40, y=0) Wait, 3.3-4.4-5.5: 3.3²+4.4²=10.89+19.36=30.25=5.5², yes! So OD length is 5.5 m.

Now, let's compute the vector for each cable:

- Cable AB: from A(0,5.60,0) to B(-4.20,0,4.20). The vector AB is B - A = (-4.20 - 0, 0 - 5.60, 4.20 - 0) = (-4.20, -5.60, 4.20)
- Cable AC: from A(0,5.60,0) to C(4.20,0,-2.40). Vector AC = (4.20 - 0, 0 - 5.60, -2.40 - 0) = (4.20, -5.60, -2.40)
- Cable AD: from A(0,5.60,0) to D(3.30,0,-4.40). Vector AD = (3.30 - 0, 0 - 5.60, -4.40 - 0) = (3.30, -5.60, -4.40)

Now, compute the length of each cable (magnitude of each vector):

- Length of AB: \( |AB| = \sqrt{(-4.20)^2 + (-5.60)^2 + 4.20^2} = \sqrt{17.64 + 31.36 + 17.64} = \sqrt{66.64} \approx 8.163 \) m? Wait, no, 4.2² + 5.6² + 4.2² = 2×17.64 + 31.36 = 35.28 + 31.36 = 66.64, sqrt≈8.163 m.

- Length of AC: \( |AC| = \sqrt{4.20^2 + (-5.60)^2 + (-2.40)^2} = \sqrt{17.64 + 31.36 + 5.76} = \sqrt{54.76} = 7.4 \) m (since 7.4²=54.76)

- Length of AD: \( |AD| = \sqrt{3.30^2 + (-5.60)^2 + (-4.40)^2} = \sqrt{10.89 + 31.36 + 19.36} = \sqrt{61.61} \approx 7.85 \) m? Wait, 3.3²=10.89, 5.6²=31.36, 4.4²=19.36; sum=10.89+31.36=42.25+19.36=61.61, sqrt≈7.85 m. Wait, but 3.3-5.6-4.4: 3.3²+4.4²=10.89+19.36=30.25=5.5², so the horizontal component (x-z) is 5.5 m, and vertical component is 5.6 m, so length of AD should be \(\sqrt{5.5^2 + 5.6^2} = \sqrt{30.25 + 31.36} = \sqrt{61.61} \approx 7.85\) m, correct.

Now, the unit vector for each cable is the vector divided by its length. The vertical component (y-component) of each tension is the tension times the y-component of the unit vector. Since the y-component of each vector is -5.60 (from A to B/C/D, the y-coordinate decreases from 5.60 to 0), the unit vector's y-component is \(\frac{-5.60}{|cable|}\). Therefore, the vertical component of each tension is \( T \times \frac{-5.60}{|cable|} \), but since the tension is pulling upward (to balance the downward force), the vertical component should be positive upward. Wait, actually, the force at A from the cable is in the direction from A to B/C

Answer:

Step1: Determine the length of each cable

First, we need to find the length of each cable (AB, AC, AD) using the distance formula in 3D. Let's assume point O is at the origin \((0,0,0)\) and point A is at \((0,5.60,0)\) (since the vertical distance from O to A is 5.60 m, and we can infer the x and z coordinates for each point:

• For cable AB: Point B is at \((-4.20, 0, 4.20)\)? Wait, no, looking at the diagram, let's get the coordinates right. Wait, the horizontal distances: For B, maybe x and z? Wait, the diagram has for B: 420 m (maybe x or z) and for C: 2.40 m (z) and 4.20 m (x)? Wait, maybe better to get the coordinates of each point:

Let’s define the coordinates:

• Point O: \((0, 0, 0)\)
• Point A: \((0, 5.60, 0)\) (since the vertical distance from O to A is 5.60 m along the y-axis)
• Point B: Let's see, the horizontal distances: From the diagram, for B, the x-distance from O is -4.20 m (assuming left), z-distance is 4.20 m? Wait, no, maybe:

Wait, the diagram shows for B: 420 m (maybe x) and 420 m (z)? Wait, no, the labels:

• For cable AB: The horizontal component (in x-z plane) distance from O to B: let's calculate the length of OB. From the diagram, OB seems to be a diagonal? Wait, no, maybe the coordinates:

Wait, let's re-express:

• Cable AB: Point B is at \((-4.20, 0, 4.20)\)? Wait, no, the horizontal distance (in x-z plane) from O to B: let's compute the length of OB. If we look at the x and z coordinates: for B, x is -4.20 m, z is 4.20 m? Then OB length is \(\sqrt{(-4.20)^2 + 4.20^2} = \sqrt{2 \times 4.20^2} = 4.20\sqrt{2}\) m. Then the length of AB is \(\sqrt{(4.20\sqrt{2})^2 + 5.60^2}\). Wait, no, AB is from A(0,5.60,0) to B(x,y,z) where B is at (x,0,z) (since it's on the horizontal plane, y=0). So the length of AB is \(\sqrt{(x - 0)^2 + (0 - 5.60)^2 + (z - 0)^2} = \sqrt{x^2 + z^2 + 5.60^2}\).

Wait, let's get the correct coordinates:

• Point B: From the diagram, the x-distance from O is -4.20 m (left), z-distance is 4.20 m (front)? Wait, no, the labels:

Wait, the problem says:

• For cable AB: The horizontal (x-z) distance from O to B: let's see the diagram, the length from O to B in x-z plane: maybe 4.20 m in x and 4.20 m in z? Wait, no, the diagram has "420 m" for B, "330 m" for D, "2.40 m" and "4.20 m" for C.

Wait, let's parse the diagram:

• Cable AB: The horizontal (x-z) component: let's say point B is at ( -4.20, 0, 4.20 ) m (so x=-4.20, z=4.20, y=0). Then the distance from O to B is \(\sqrt{(-4.20)^2 + 4.20^2} = \sqrt{2 \times 4.20^2} = 4.20\sqrt{2}\) m. Then the length of AB is \(\sqrt{(4.20\sqrt{2})^2 + 5.60^2}\). Let's calculate that:

\( (4.20\sqrt{2})^2 = 4.20^2 \times 2 = 17.64 \times 2 = 35.28 \)

\( 5.60^2 = 31.36 \)

So AB length: \(\sqrt{35.28 + 31.36} = \sqrt{66.64} \approx 8.163\) m? Wait, no, that can't be. Wait, maybe the vertical distance is 5.60 m (along y-axis), and the horizontal distance (x-z) for AB is 7.0 m? Wait, 4.20 and 5.60? Wait, 4.2-5.6-7.0 is a 3-4-5 triangle scaled by 1.4: 4.2=3×1.4, 5.6=4×1.4, so 7.0=5×1.4. Ah! That makes sense. So maybe the horizontal distance (x-z) for AB is 7.0 m? Wait, 4.20 and 5.60? Wait, 4.20 is 3×1.4, 5.60 is 4×1.4, so the hypotenuse (horizontal) would be 5×1.4=7.0 m. Then the length of AB is \(\sqrt{7.0^2 + 5.60^2}\)? No, wait, AB is from A(0,5.60,0) to B(x,0,z), so the vertical component is 5.60 m (along y-axis), and the horizontal component (x-z) is the distance from O to B, which is \(\sqrt{x^2 + z^2}\). If the horizontal component is 7.0 m (since 4.2 and 5.6 are 3:4, so hypotenuse 5×1.4=7), then AB length is \(\sqrt{7.0^2 + 5.60^2}\)? Wait, no, 5.60 is the vertical distance (y-axis), so the length of AB is \(\sqrt{(x - 0)^2 + (0 - 5.60)^2 + (z - 0)^2} = \sqrt{x^2 + z^2 + 5.60^2}\). If x^2 + z^2 = 7.0^2 = 49, then AB length is \(\sqrt{49 + 31.36} = \sqrt{80.36} \approx 8.964\) m? Wait, no, I think I messed up the coordinates.

Wait, let's start over. Let's define the coordinates properly:

• Let O be at (0, 0, 0).
• Point A is at (0, 5.60, 0) (since the vertical distance from O to A is 5.60 m along the y-axis).
• Point B: Let's look at the diagram, the horizontal (x-z) distance from O to B: the x-coordinate is -4.20 m (left), z-coordinate is 4.20 m (front)? Wait, no, the diagram shows "420 m" for B, "330 m" for D, "2.40 m" and "4.20 m" for C.

Wait, maybe:

• Cable AB: The horizontal (x-z) component: x = -4.20 m, z = 4.20 m (so OB vector is (-4.20, 0, 4.20)). Then the length of OB is \(\sqrt{(-4.20)^2 + 4.20^2} = \sqrt{2 \times 4.20^2} = 4.20\sqrt{2} \approx 5.94\) m. Then the length of AB is \(\sqrt{(4.20\sqrt{2})^2 + 5.60^2} = \sqrt{35.28 + 31.36} = \sqrt{66.64} \approx 8.16\) m.

• Cable AC: Point C is at (4.20, 0, -2.40) (x=4.20, z=-2.40). Then OB (OC) length is \(\sqrt{4.20^2 + (-2.40)^2} = \sqrt{17.64 + 5.76} = \sqrt{23.4} \approx 4.837\) m. Then AC length is \(\sqrt{4.837^2 + 5.60^2} \approx \sqrt{23.4 + 31.36} = \sqrt{54.76} \approx 7.4\) m? Wait, no, 4.20^2 + 2.40^2 = 17.64 + 5.76 = 23.4, square root is ~4.837, then 4.837^2 + 5.60^2 = 23.4 + 31.36 = 54.76, square root is 7.4 m (since 7.4^2 = 54.76). Ah, that's a nice number.

• Cable AD: Point D is at (3.30, 0, -4.40)? Wait, no, the diagram shows "330 m" and "5.60 m"? Wait, no, the horizontal (x-z) for D: x=3.30, z=-4.40? Wait, 3.3-4.4-5.5 is a 3-4-5 triangle scaled by 1.1: 3.3=3×1.1, 4.4=4×1.1, so 5.5=5×1.1. Then the length of OD is 5.5 m, and AD length is \(\sqrt{5.5^2 + 5.60^2}\)? Wait, no, 5.5^2 = 30.25, 5.60^2=31.36, so AD length is \(\sqrt{30.25 + 31.36} = \sqrt{61.61} \approx 7.85\) m? No, that doesn't fit. Wait, maybe the vertical distance is 5.60 m, and the horizontal distances (x-z) for each cable are:

• AB: horizontal distance (OD) = 7.0 m (since 4.2-5.6-7.0 is 3-4-5×1.4), so AB length = \(\sqrt{7.0^2 + 5.60^2}\)? Wait, 7.0^2=49, 5.60^2=31.36, sum=80.36, sqrt=8.964≈9.0 m? No, 7-5.6-8.96 is 5-4-6.4, which is a similar triangle.

Wait, maybe the key is that the vertical force at A is 800 N upward (wait, no, the balloon exerts a vertical force at A, probably downward? Wait, the balloon is tethered, so the force at A is downward (weight), and the cables pull upward. So we have three cables, each with tension T_AB, T_AC, T_AD, and the vertical component of each tension must sum to 800 N (since the system is in equilibrium, vertical forces balance: sum of vertical components of tensions = 800 N downward? Wait, no: if the balloon exerts a 800 N vertical force downward at A, then the sum of the vertical components of the cable tensions must be 800 N upward.

To find the vertical component of each tension, we need the unit vector in the direction of each cable, then take the y-component (since the vertical direction is y).

First, let's find the coordinates of each point:

• Point A: (0, 5.60, 0)
• Point B: Let's assume from the diagram, the horizontal (x-z) coordinates of B are (-4.20, 0, 4.20) (so x=-4.20, z=4.20, y=0)
• Point C: (4.20, 0, -2.40) (x=4.20, z=-2.40, y=0)
• Point D: (3.30, 0, -4.40) (x=3.30, z=-4.40, y=0) Wait, 3.3-4.4-5.5: 3.3²+4.4²=10.89+19.36=30.25=5.5², yes! So OD length is 5.5 m.

Now, let's compute the vector for each cable:

• Cable AB: from A(0,5.60,0) to B(-4.20,0,4.20). The vector AB is B - A = (-4.20 - 0, 0 - 5.60, 4.20 - 0) = (-4.20, -5.60, 4.20)
• Cable AC: from A(0,5.60,0) to C(4.20,0,-2.40). Vector AC = (4.20 - 0, 0 - 5.60, -2.40 - 0) = (4.20, -5.60, -2.40)
• Cable AD: from A(0,5.60,0) to D(3.30,0,-4.40). Vector AD = (3.30 - 0, 0 - 5.60, -4.40 - 0) = (3.30, -5.60, -4.40)

Now, compute the length of each cable (magnitude of each vector):

• Length of AB: \( |AB| = \sqrt{(-4.20)^2 + (-5.60)^2 + 4.20^2} = \sqrt{17.64 + 31.36 + 17.64} = \sqrt{66.64} \approx 8.163 \) m? Wait, no, 4.2² + 5.6² + 4.2² = 2×17.64 + 31.36 = 35.28 + 31.36 = 66.64, sqrt≈8.163 m.

• Length of AC: \( |AC| = \sqrt{4.20^2 + (-5.60)^2 + (-2.40)^2} = \sqrt{17.64 + 31.36 + 5.76} = \sqrt{54.76} = 7.4 \) m (since 7.4²=54.76)

• Length of AD: \( |AD| = \sqrt{3.30^2 + (-5.60)^2 + (-4.40)^2} = \sqrt{10.89 + 31.36 + 19.36} = \sqrt{61.61} \approx 7.85 \) m? Wait, 3.3²=10.89, 5.6²=31.36, 4.4²=19.36; sum=10.89+31.36=42.25+19.36=61.61, sqrt≈7.85 m. Wait, but 3.3-5.6-4.4: 3.3²+4.4²=10.89+19.36=30.25=5.5², so the horizontal component (x-z) is 5.5 m, and vertical component is 5.6 m, so length of AD should be \(\sqrt{5.5^2 + 5.6^2} = \sqrt{30.25 + 31.36} = \sqrt{61.61} \approx 7.85\) m, correct.

Now, the unit vector for each cable is the vector divided by its length. The vertical component (y-component) of each tension is the tension times the y-component of the unit vector. Since the y-component of each vector is -5.60 (from A to B/C/D, the y-coordinate decreases from 5.60 to 0), the unit vector's y-component is \(\frac{-5.60}{|cable|}\). Therefore, the vertical component of each tension is \( T \times \frac{-5.60}{|cable|} \), but since the tension is pulling upward (to balance the downward force), the vertical component should be positive upward. Wait, actually, the force at A from the cable is in the direction from A to B/C