Question

problem 3.30 consider the system in (figure 1). part a determine the maximum weight of the bucket that the wire system can support so that no single wire develops a tension exceeding 100 lb. express your answer with the appropriate units.

Explanation:

Step1: Analyze Joint B

Let's assume the tension in wire BC is \( T_{BC} \) and in wire BE is \( T_{BE} \). At joint B, the vertical equilibrium: \( T_{BC}\sin30^{\circ}=T_{BE}\sin\theta \), where \( \theta \) is the angle for BE. From the 3-4-5 triangle, \( \sin\theta=\frac{3}{5} \), \( \cos\theta=\frac{4}{5} \).

Step2: Analyze Joint E

At joint E, the vertical component from BE is \( T_{BE}\sin\theta \), and this should balance the weight \( W \). Also, the horizontal component from BE is \( T_{BE}\cos\theta \), and the horizontal component from the bucket wire (let's say \( T_{bucket} \)) is balanced. But first, find the maximum tension in each wire. The wire with 3-4-5 triangle: the tension in BE, if we consider the angles, let's first check the tension limits. The maximum tension in any wire is 100 lb. Let's check which wire reaches 100 lb first.

For wire BC: \( T_{BC} \leq 100 \) lb. Then vertical force from BC: \( T_{BC}\sin30^{\circ}=100\times0.5 = 50 \) lb.

For wire BE: The vertical component from BE is \( T_{BE}\sin\theta = T_{BE}\times\frac{3}{5} \). The horizontal component is \( T_{BE}\times\frac{4}{5} \).

Now, at joint E, the vertical force is \( W \), and the horizontal force is balanced by the bucket's horizontal component (but since the bucket is vertical, maybe the horizontal forces at E are from BE and the other wire? Wait, maybe better to do equilibrium at each joint.

Wait, the 3-4-5 triangle: so the length of BE's vertical component is 3, horizontal is 4, hypotenuse 5. So the angle for BE with horizontal is \( \arcsin(\frac{3}{5}) \), so vertical component is \( \frac{3}{5}T_{BE} \), horizontal is \( \frac{4}{5}T_{BE} \).

At joint B: vertical equilibrium: \( T_{BC}\sin30^{\circ}=T_{BE}\sin\theta \) (where \( \theta \) is the angle for BE, \( \sin\theta=\frac{3}{5} \)), so \( T_{BC}\times0.5 = T_{BE}\times\frac{3}{5} \), so \( T_{BC}=\frac{6}{5}T_{BE} \). So if \( T_{BE} \) is at maximum, \( T_{BC} \) would be larger. So the limiting factor is \( T_{BC} \) or \( T_{BE} \)? Wait, if \( T_{BC} \) max is 100, then \( T_{BE}=\frac{5}{6}\times100\approx83.33 \) lb. If \( T_{BE} \) max is 100, then \( T_{BC}=\frac{6}{5}\times100 = 120 \) lb, which exceeds 100, so \( T_{BC} \) can't be 120. So \( T_{BC} \) is limited to 100 lb, so \( T_{BE}=\frac{5}{6}\times100\approx83.33 \) lb.

Now, at joint E: the vertical force is \( W \), which is balanced by the vertical component of \( T_{BE} \) and maybe another wire? Wait, the bucket is hanging from E, so the vertical force at E is \( W \), and the vertical component from BE is \( T_{BE}\sin\theta = 83.33\times\frac{3}{5}=50 \) lb? Wait, no, maybe I messed up the joints. Wait, the figure: A is a pulley, B is a joint, E is another joint, with a 3-4-5 triangle between B and E, and angles of 30 degrees for BC and ED? Wait, maybe the correct approach is:

Let's denote:

- Tension in AB: let's say it's vertical? No, AB is connected to B, and BC is at 30 degrees, BE is at an angle with 3-4-5 triangle (so 3-4-5, so opposite 3, adjacent 4, hypotenuse 5, so angle \( \alpha \) where \( \sin\alpha = 3/5 \), \( \cos\alpha = 4/5 \)).

At joint B:

Forces:

- Vertical: \( T_{AB} = T_{BC}\sin30^{\circ} + T_{BE}\sin\alpha \)

- Horizontal: \( T_{BC}\cos30^{\circ} = T_{BE}\cos\alpha \)

At joint E:

Forces:

- Vertical: \( W = T_{BE}\sin\alpha + T_{ED}\sin30^{\circ} \) (assuming ED is at 30 degrees, similar to BC)

- Horizontal: \( T_{BE}\cos\alpha = T_{ED}\cos30^{\circ} \)

Wait, maybe the system is symmetric? So \( T_{BC}=T_{ED} \) and \( T_{BE} \) is the same on both sides? Wait, the 3-4-5 triangle: so BE has length 5, vertical component 3, horizontal 4. So \( \sin\alpha = 3/5 \), \( \cos\alpha = 4/5 \).

From horizontal equilibrium at B: \( T_{BC}\cos30^{\circ} = T_{BE}\cos\alpha \)

So \( T_{BE} = T_{BC}\frac{\cos30^{\circ}}{\cos\alpha} = T_{BC}\frac{\sqrt{3}/2}{4/5} = T_{BC}\frac{5\sqrt{3}}{8} \approx T_{BC}\times1.0825 \)

So \( T_{BE} \) is larger than \( T_{BC} \) if \( T_{BC} \) is positive. So if \( T_{BE} \) max is 100 lb, then \( T_{BC}=100\times\frac{8}{5\sqrt{3}}\approx100\times0.9238\approx92.38 \) lb, which is less than 100, so \( T_{BE} \) would reach 100 first? Wait, no, \( T_{BE}=T_{BC}\times1.0825 \), so if \( T_{BC}=100 \), \( T_{BE}=108.25 \), which exceeds 100. So \( T_{BE} \) can't exceed 100, so \( T_{BC}=100 / 1.0825\approx92.38 \) lb.

Now, at joint E: vertical equilibrium: \( W = T_{BE}\sin\alpha + T_{ED}\sin30^{\circ} \). If symmetric, \( T_{ED}=T_{BC} \), so \( W = T_{BE}\times\frac{3}{5} + T_{BC}\times0.5 \)

Substitute \( T_{BE}=100 \), \( T_{BC}=92.38 \):

\( W = 100\times\frac{3}{5} + 92.38\times0.5 = 60 + 46.19 = 106.19 \)? No, that can't be. Wait, maybe I got the angles wrong. Wait, the 3-4-5 triangle: the vertical side is 3, horizontal is 4, so the angle with the horizontal is \( \arctan(3/4) \approx 36.87^\circ \). The other wires (BC and ED) are at 30 degrees. So let's redo:

At joint B:

Horizontal: \( T_{BC}\cos30^\circ = T_{BE}\cos(36.87^\circ) \)

\( \cos30^\circ \approx 0.8660 \), \( \cos36.87^\circ = 4/5 = 0.8 \)

So \( T_{BE} = T_{BC}\times\frac{0.8660}{0.8} \approx T_{BC}\times1.0825 \)

Vertical: \( T_{AB} = T_{BC}\sin30^\circ + T_{BE}\sin36.87^\circ \)

\( \sin30^\circ = 0.5 \), \( \sin36.87^\circ = 3/5 = 0.6 \)

So \( T_{AB} = 0.5T_{BC} + 0.6T_{BE} = 0.5T_{BC} + 0.6\times1.0825T_{BC} = 0.5T_{BC} + 0.6495T_{BC} = 1.1495T_{BC} \)

At joint E:

Horizontal: \( T_{BE}\cos36.87^\circ = T_{ED}\cos30^\circ \)

So \( T_{ED} = T_{BE}\times\frac{0.8}{0.8660} \approx T_{BE}\times0.9238 \approx T_{BC}\times1.0825\times0.9238 \approx T_{BC} \) (since 1.0825×0.9238≈1), so symmetric, \( T_{ED}=T_{BC} \)

Vertical: \( W = T_{BE}\sin36.87^\circ + T_{ED}\sin30^\circ = 0.6T_{BE} + 0.5T_{BC} \)

Now, the maximum tension in any wire: \( T_{AB} \), \( T_{BC} \), \( T_{BE} \), \( T_{ED} \). We saw \( T_{BE} \approx 1.0825T_{BC} \), \( T_{AB} \approx 1.1495T_{BC} \). So \( T_{AB} \) is the largest? Wait, no, if \( T_{BC}=100 \), \( T_{BE}=108.25 \) (exceeds 100), \( T_{AB}=114.95 \) (exceeds 100). So the limiting wire is \( T_{BE} \) or \( T_{AB} \)? Wait, maybe the problem is that the wires BC and ED have maximum tension 100 lb, and BE and AB? Wait, the problem says "no single wire develops a tension exceeding 100 lb". So we need to find which wire has the maximum tension for a given W, then set that tension to 100 lb and solve for W.

Let's denote:

Let \( T_{BC} = T_{ED} = T_1 \) (tension in BC and ED)

\( T_{BE} = T_2 \) (tension in BE)

\( T_{AB} = T_3 \) (tension in AB)

From joint B horizontal: \( T_1\cos30^\circ = T_2\cos\alpha \), where \( \alpha = \arctan(3/4) \), so \( \cos\alpha = 4/5 \), \( \sin\alpha = 3/5 \)

Thus, \( T_2 = T_1 \times \frac{\cos30^\circ}{4/5} = T_1 \times \frac{5\sqrt{3}}{8} \approx 1.0825T_1 \)

From joint B vertical: \( T_3 = T_1\sin30^\circ + T_2\sin\alpha = 0.5T_1 + T_2\times\frac{3}{5} \)

Substitute \( T_2 \):

\( T_3 = 0.5T_1 + (1.0825T_1)\times0.6 = 0.5T_1 + 0.6495T_1 = 1.1495T_1 \)

From joint E vertical: \( W = T_2\sin\alpha + T_1\sin30^\circ = 0.6T_2 + 0.5T_1 \)

Substitute \( T_2 \):

\( W = 0.6\times1.0825T_1 + 0.5T_1 = 0.6495T_1 + 0.5T_1 = 1.1495T_1 \)

Now, let's check the tensions:

- \( T_2 \approx 1.0825T_1 \)

- \( T_3 \approx 1.1495T_1 \)

So \( T_3 > T_2 > T_1 \). So the maximum tension is in AB (\( T_3 \)). Wait, but maybe the wires are BC, BE, ED, AB. Wait, maybe the 3-4-5 triangle is between B and E, so BE is a wire, and BC and ED are at 30 degrees. Wait, maybe I made a mistake in the joint E. Let's look at the figure again: E is connected to B (3-4-5 triangle), to D (30 degrees), and to the bucket (vertical). So at E, forces:

- Vertical: bucket weight \( W \) downward, \( T_{BE}\sin\alpha \) upward (from BE), \( T_{ED}\sin30^\circ \) upward (from ED)

- Horizontal: \( T_{BE}\cos\alpha \) to the right, \( T_{ED}\cos30^\circ \) to the left

So horizontal equilibrium: \( T_{BE}\cos\alpha = T_{ED}\cos30^\circ \) --> \( T_{ED} = T_{BE}\frac{\cos\alpha}{\cos30^\circ} = T_{BE}\frac{4/5}{\sqrt{3}/2} = T_{BE}\frac{8}{5\sqrt{3}} \approx 0.9238T_{BE} \)

Vertical equilibrium: \( W = T_{BE}\sin\alpha + T_{ED}\sin30^\circ = T_{BE}\times\frac{3}{5} + T_{ED}\times0.5 \)

Substitute \( T_{ED} \):

\( W = 0.6T_{BE} + 0.5\times0.9238T_{BE} = 0.6T_{BE} + 0.4619T_{BE} = 1.0619T_{BE} \)

At joint B:

Forces:

- Vertical: \( T_{AB} = T_{BC}\sin30^\circ + T_{BE}\sin\alpha = 0.5T_{BC} + 0.6T_{BE} \)

- Horizontal: \( T_{BC}\cos30^\circ = T_{BE}\cos\alpha \) --> \( T_{BC} = T_{BE}\frac{\cos\alpha}{\cos30^\circ} = 0.9238T_{BE} \) (same as \( T_{ED} \))

So \( T_{AB} = 0.5\times0.9238T_{BE} + 0.6T_{BE} = 0.4619T_{BE} + 0.6T_{BE} = 1.0619T_{BE} \)

Now, the tensions:

- \( T_{BC} = T_{ED} \approx 0.9238T_{BE} \)

- \( T_{AB} = W \approx 1.0619T_{BE} \)

So the maximum tension is in AB and W (since \( W = T_{AB} \))? Wait, no, \( W = 1.0619T_{BE} \), and \( T_{AB} = W \), \( T_{BC} = 0.9238T_{BE} \), \( T_{BE} \) is a wire. So the wires are AB, BC, BE, ED. So we need to check which wire has tension ≤100 lb.

Case 1: \( T_{BE} = 100 \) lb. Then \( T_{BC} = 0.9238×100 = 92.38 \) lb, \( W = 1.0619×100 = 106.19 \) lb, \( T_{AB} = 106.19 \) lb. But is \( T_{AB} \) a wire? If AB is a wire, then its tension is 106.19 lb, which exceeds 100? Wait, the problem says "no single wire develops a tension exceeding 100 lb". So if AB is a wire, then \( T_{AB} \leq 100 \), so \( W \leq 100 \), then \( T_{BE} = W / 1.0619 \approx 94.17 \) lb, \( T_{BC} = 0.9238×94.17≈87.0 \) lb. But that can't be right. Wait, maybe the 3-4-5 triangle is such that the vertical side is 4 and horizontal is 3? No, 3-4-5 is opposite 3, adjacent 4.

Wait, maybe the correct

Answer:

Step1: Analyze Joint B

Let's assume the tension in wire BC is \( T_{BC} \) and in wire BE is \( T_{BE} \). At joint B, the vertical equilibrium: \( T_{BC}\sin30^{\circ}=T_{BE}\sin\theta \), where \( \theta \) is the angle for BE. From the 3-4-5 triangle, \( \sin\theta=\frac{3}{5} \), \( \cos\theta=\frac{4}{5} \).

Step2: Analyze Joint E

At joint E, the vertical component from BE is \( T_{BE}\sin\theta \), and this should balance the weight \( W \). Also, the horizontal component from BE is \( T_{BE}\cos\theta \), and the horizontal component from the bucket wire (let's say \( T_{bucket} \)) is balanced. But first, find the maximum tension in each wire. The wire with 3-4-5 triangle: the tension in BE, if we consider the angles, let's first check the tension limits. The maximum tension in any wire is 100 lb. Let's check which wire reaches 100 lb first.

For wire BC: \( T_{BC} \leq 100 \) lb. Then vertical force from BC: \( T_{BC}\sin30^{\circ}=100\times0.5 = 50 \) lb.

For wire BE: The vertical component from BE is \( T_{BE}\sin\theta = T_{BE}\times\frac{3}{5} \). The horizontal component is \( T_{BE}\times\frac{4}{5} \).

Now, at joint E, the vertical force is \( W \), and the horizontal force is balanced by the bucket's horizontal component (but since the bucket is vertical, maybe the horizontal forces at E are from BE and the other wire? Wait, maybe better to do equilibrium at each joint.

Wait, the 3-4-5 triangle: so the length of BE's vertical component is 3, horizontal is 4, hypotenuse 5. So the angle for BE with horizontal is \( \arcsin(\frac{3}{5}) \), so vertical component is \( \frac{3}{5}T_{BE} \), horizontal is \( \frac{4}{5}T_{BE} \).

At joint B: vertical equilibrium: \( T_{BC}\sin30^{\circ}=T_{BE}\sin\theta \) (where \( \theta \) is the angle for BE, \( \sin\theta=\frac{3}{5} \)), so \( T_{BC}\times0.5 = T_{BE}\times\frac{3}{5} \), so \( T_{BC}=\frac{6}{5}T_{BE} \). So if \( T_{BE} \) is at maximum, \( T_{BC} \) would be larger. So the limiting factor is \( T_{BC} \) or \( T_{BE} \)? Wait, if \( T_{BC} \) max is 100, then \( T_{BE}=\frac{5}{6}\times100\approx83.33 \) lb. If \( T_{BE} \) max is 100, then \( T_{BC}=\frac{6}{5}\times100 = 120 \) lb, which exceeds 100, so \( T_{BC} \) can't be 120. So \( T_{BC} \) is limited to 100 lb, so \( T_{BE}=\frac{5}{6}\times100\approx83.33 \) lb.

Now, at joint E: the vertical force is \( W \), which is balanced by the vertical component of \( T_{BE} \) and maybe another wire? Wait, the bucket is hanging from E, so the vertical force at E is \( W \), and the vertical component from BE is \( T_{BE}\sin\theta = 83.33\times\frac{3}{5}=50 \) lb? Wait, no, maybe I messed up the joints. Wait, the figure: A is a pulley, B is a joint, E is another joint, with a 3-4-5 triangle between B and E, and angles of 30 degrees for BC and ED? Wait, maybe the correct approach is:

Let's denote:

• Tension in AB: let's say it's vertical? No, AB is connected to B, and BC is at 30 degrees, BE is at an angle with 3-4-5 triangle (so 3-4-5, so opposite 3, adjacent 4, hypotenuse 5, so angle \( \alpha \) where \( \sin\alpha = 3/5 \), \( \cos\alpha = 4/5 \)).

At joint B:

Forces:

• Vertical: \( T_{AB} = T_{BC}\sin30^{\circ} + T_{BE}\sin\alpha \)

• Horizontal: \( T_{BC}\cos30^{\circ} = T_{BE}\cos\alpha \)

At joint E:

Forces:

• Vertical: \( W = T_{BE}\sin\alpha + T_{ED}\sin30^{\circ} \) (assuming ED is at 30 degrees, similar to BC)

• Horizontal: \( T_{BE}\cos\alpha = T_{ED}\cos30^{\circ} \)

Wait, maybe the system is symmetric? So \( T_{BC}=T_{ED} \) and \( T_{BE} \) is the same on both sides? Wait, the 3-4-5 triangle: so BE has length 5, vertical component 3, horizontal 4. So \( \sin\alpha = 3/5 \), \( \cos\alpha = 4/5 \).

From horizontal equilibrium at B: \( T_{BC}\cos30^{\circ} = T_{BE}\cos\alpha \)

So \( T_{BE} = T_{BC}\frac{\cos30^{\circ}}{\cos\alpha} = T_{BC}\frac{\sqrt{3}/2}{4/5} = T_{BC}\frac{5\sqrt{3}}{8} \approx T_{BC}\times1.0825 \)

So \( T_{BE} \) is larger than \( T_{BC} \) if \( T_{BC} \) is positive. So if \( T_{BE} \) max is 100 lb, then \( T_{BC}=100\times\frac{8}{5\sqrt{3}}\approx100\times0.9238\approx92.38 \) lb, which is less than 100, so \( T_{BE} \) would reach 100 first? Wait, no, \( T_{BE}=T_{BC}\times1.0825 \), so if \( T_{BC}=100 \), \( T_{BE}=108.25 \), which exceeds 100. So \( T_{BE} \) can't exceed 100, so \( T_{BC}=100 / 1.0825\approx92.38 \) lb.

Now, at joint E: vertical equilibrium: \( W = T_{BE}\sin\alpha + T_{ED}\sin30^{\circ} \). If symmetric, \( T_{ED}=T_{BC} \), so \( W = T_{BE}\times\frac{3}{5} + T_{BC}\times0.5 \)

Substitute \( T_{BE}=100 \), \( T_{BC}=92.38 \):

\( W = 100\times\frac{3}{5} + 92.38\times0.5 = 60 + 46.19 = 106.19 \)? No, that can't be. Wait, maybe I got the angles wrong. Wait, the 3-4-5 triangle: the vertical side is 3, horizontal is 4, so the angle with the horizontal is \( \arctan(3/4) \approx 36.87^\circ \). The other wires (BC and ED) are at 30 degrees. So let's redo:

At joint B:

Horizontal: \( T_{BC}\cos30^\circ = T_{BE}\cos(36.87^\circ) \)

\( \cos30^\circ \approx 0.8660 \), \( \cos36.87^\circ = 4/5 = 0.8 \)

So \( T_{BE} = T_{BC}\times\frac{0.8660}{0.8} \approx T_{BC}\times1.0825 \)

Vertical: \( T_{AB} = T_{BC}\sin30^\circ + T_{BE}\sin36.87^\circ \)

\( \sin30^\circ = 0.5 \), \( \sin36.87^\circ = 3/5 = 0.6 \)

So \( T_{AB} = 0.5T_{BC} + 0.6T_{BE} = 0.5T_{BC} + 0.6\times1.0825T_{BC} = 0.5T_{BC} + 0.6495T_{BC} = 1.1495T_{BC} \)

At joint E:

Horizontal: \( T_{BE}\cos36.87^\circ = T_{ED}\cos30^\circ \)

So \( T_{ED} = T_{BE}\times\frac{0.8}{0.8660} \approx T_{BE}\times0.9238 \approx T_{BC}\times1.0825\times0.9238 \approx T_{BC} \) (since 1.0825×0.9238≈1), so symmetric, \( T_{ED}=T_{BC} \)

Vertical: \( W = T_{BE}\sin36.87^\circ + T_{ED}\sin30^\circ = 0.6T_{BE} + 0.5T_{BC} \)

Now, the maximum tension in any wire: \( T_{AB} \), \( T_{BC} \), \( T_{BE} \), \( T_{ED} \). We saw \( T_{BE} \approx 1.0825T_{BC} \), \( T_{AB} \approx 1.1495T_{BC} \). So \( T_{AB} \) is the largest? Wait, no, if \( T_{BC}=100 \), \( T_{BE}=108.25 \) (exceeds 100), \( T_{AB}=114.95 \) (exceeds 100). So the limiting wire is \( T_{BE} \) or \( T_{AB} \)? Wait, maybe the problem is that the wires BC and ED have maximum tension 100 lb, and BE and AB? Wait, the problem says "no single wire develops a tension exceeding 100 lb". So we need to find which wire has the maximum tension for a given W, then set that tension to 100 lb and solve for W.

Let's denote:

Let \( T_{BC} = T_{ED} = T_1 \) (tension in BC and ED)

\( T_{BE} = T_2 \) (tension in BE)

\( T_{AB} = T_3 \) (tension in AB)

From joint B horizontal: \( T_1\cos30^\circ = T_2\cos\alpha \), where \( \alpha = \arctan(3/4) \), so \( \cos\alpha = 4/5 \), \( \sin\alpha = 3/5 \)

Thus, \( T_2 = T_1 \times \frac{\cos30^\circ}{4/5} = T_1 \times \frac{5\sqrt{3}}{8} \approx 1.0825T_1 \)

From joint B vertical: \( T_3 = T_1\sin30^\circ + T_2\sin\alpha = 0.5T_1 + T_2\times\frac{3}{5} \)

Substitute \( T_2 \):

\( T_3 = 0.5T_1 + (1.0825T_1)\times0.6 = 0.5T_1 + 0.6495T_1 = 1.1495T_1 \)

From joint E vertical: \( W = T_2\sin\alpha + T_1\sin30^\circ = 0.6T_2 + 0.5T_1 \)

Substitute \( T_2 \):

\( W = 0.6\times1.0825T_1 + 0.5T_1 = 0.6495T_1 + 0.5T_1 = 1.1495T_1 \)

Now, let's check the tensions:

• \( T_2 \approx 1.0825T_1 \)

• \( T_3 \approx 1.1495T_1 \)

So \( T_3 > T_2 > T_1 \). So the maximum tension is in AB (\( T_3 \)). Wait, but maybe the wires are BC, BE, ED, AB. Wait, maybe the 3-4-5 triangle is between B and E, so BE is a wire, and BC and ED are at 30 degrees. Wait, maybe I made a mistake in the joint E. Let's look at the figure again: E is connected to B (3-4-5 triangle), to D (30 degrees), and to the bucket (vertical). So at E, forces:

• Vertical: bucket weight \( W \) downward, \( T_{BE}\sin\alpha \) upward (from BE), \( T_{ED}\sin30^\circ \) upward (from ED)

• Horizontal: \( T_{BE}\cos\alpha \) to the right, \( T_{ED}\cos30^\circ \) to the left

So horizontal equilibrium: \( T_{BE}\cos\alpha = T_{ED}\cos30^\circ \) --> \( T_{ED} = T_{BE}\frac{\cos\alpha}{\cos30^\circ} = T_{BE}\frac{4/5}{\sqrt{3}/2} = T_{BE}\frac{8}{5\sqrt{3}} \approx 0.9238T_{BE} \)

Vertical equilibrium: \( W = T_{BE}\sin\alpha + T_{ED}\sin30^\circ = T_{BE}\times\frac{3}{5} + T_{ED}\times0.5 \)

Substitute \( T_{ED} \):

\( W = 0.6T_{BE} + 0.5\times0.9238T_{BE} = 0.6T_{BE} + 0.4619T_{BE} = 1.0619T_{BE} \)

At joint B:

Forces:

• Vertical: \( T_{AB} = T_{BC}\sin30^\circ + T_{BE}\sin\alpha = 0.5T_{BC} + 0.6T_{BE} \)

• Horizontal: \( T_{BC}\cos30^\circ = T_{BE}\cos\alpha \) --> \( T_{BC} = T_{BE}\frac{\cos\alpha}{\cos30^\circ} = 0.9238T_{BE} \) (same as \( T_{ED} \))

So \( T_{AB} = 0.5\times0.9238T_{BE} + 0.6T_{BE} = 0.4619T_{BE} + 0.6T_{BE} = 1.0619T_{BE} \)

Now, the tensions:

• \( T_{BC} = T_{ED} \approx 0.9238T_{BE} \)

• \( T_{AB} = W \approx 1.0619T_{BE} \)

So the maximum tension is in AB and W (since \( W = T_{AB} \))? Wait, no, \( W = 1.0619T_{BE} \), and \( T_{AB} = W \), \( T_{BC} = 0.9238T_{BE} \), \( T_{BE} \) is a wire. So the wires are AB, BC, BE, ED. So we need to check which wire has tension ≤100 lb.

Case 1: \( T_{BE} = 100 \) lb. Then \( T_{BC} = 0.9238×100 = 92.38 \) lb, \( W = 1.0619×100 = 106.19 \) lb, \( T_{AB} = 106.19 \) lb. But is \( T_{AB} \) a wire? If AB is a wire, then its tension is 106.19 lb, which exceeds 100? Wait, the problem says "no single wire develops a tension exceeding 100 lb". So if AB is a wire, then \( T_{AB} \leq 100 \), so \( W \leq 100 \), then \( T_{BE} = W / 1.0619 \approx 94.17 \) lb, \( T_{BC} = 0.9238×94.17≈87.0 \) lb. But that can't be right. Wait, maybe the 3-4-5 triangle is such that the vertical side is 4 and horizontal is 3? No, 3-4-5 is opposite 3, adjacent 4.

Wait, maybe the correct