Question

a frame hanging on a wall is held by two cables. the tension in each cable is 30 n, and the cables make an angle of 45° with the horizontal, as shown in the picture. what is the weight of the frame? a 10.2 n b 21.2 n c 32.4 n d 42.4 n

Explanation:

Step1: Analyze Vertical Forces

The frame is in equilibrium, so vertical forces sum to zero. Each cable has tension \( T = 30\,\text{N} \) at \( 45^\circ \) to horizontal. The vertical component of each tension is \( T_y = T\sin(45^\circ) \).
\[ T_y = 30 \times \sin(45^\circ) = 30 \times \frac{\sqrt{2}}{2} \approx 30 \times 0.7071 \approx 21.21\,\text{N} \]

Step2: Total Vertical Force (Weight)

There are two cables, so total vertical force (weight \( W \)) is sum of both vertical components.
\[ W = 2 \times T_y = 2 \times 21.21 \approx 42.42\,\text{N} \] Wait, no—wait, maybe the diagram has one cable? Wait, the problem says "two cables" but the diagram shows one? Wait, re-reading: "A frame hanging on a wall is held by two cables. The tension in each cable is 30 N, and the cables make an angle of 45° with the horizontal..." So each cable has \( T = 30\,\text{N} \), angle \( 45^\circ \). So vertical component per cable: \( T\sin(45^\circ) \), two cables: \( 2 \times 30 \times \sin(45^\circ) \)? Wait, no—wait, maybe the diagram is a single cable? Wait, the user's diagram: "30N" and "45°", maybe it's one cable? Wait, the problem says two cables. Wait, maybe the diagram is a free-body diagram with two tensions? Wait, no, the options are A:10.2, B:21.2, C:32.4, D:42.4. Wait, if it's one cable, vertical component is \( 30\sin(45^\circ) \approx 21.2\), but two cables? Wait, no—maybe the diagram is for one cable, but the problem says two? Wait, no, let's check the options. Option B is 21.2, D is 42.4. Wait, maybe the diagram has two cables, each at 45°, so each has vertical component \( 30\sin(45) \), so total weight is sum. Wait, no—equilibrium: weight equals total upward force. If two cables, each with \( T = 30\,\text{N} \), angle \( 45^\circ \), then vertical force from each is \( 30\sin(45) \), so total upward force is \( 2 \times 30\sin(45) \approx 42.4\). But option D is 42.4. Wait, but maybe the diagram is one cable? Wait, the problem statement: "the cables make an angle of 45° with the horizontal, as shown in the picture". The picture shows a triangle with 30N, 45°, and \( F_y \). Maybe it's one cable? Wait, then weight would be \( 30\sin(45) \approx 21.2\), which is option B. Wait, confusion here. Wait, the problem says "two cables", tension in each is 30N. So each cable: tension 30N, angle 45° to horizontal. So vertical component per cable: \( 30\sin(45°) \approx 21.2\,\text{N} \). If two cables, total upward force is \( 2 \times 21.2 = 42.4\,\text{N} \), which is option D. But the diagram shows one cable? Wait, maybe the problem has a typo, or the diagram is for one cable. Wait, the options: B is 21.2, D is 42.4. Let's recalculate:

If one cable: \( W = T\sin(45°) = 30 \times \frac{\sqrt{2}}{2} \approx 21.2\,\text{N} \) (option B).

If two cables: \( W = 2 \times 30 \times \sin(45°) \approx 42.4\,\text{N} \) (option D).

But the problem says "two cables", tension in each is 30N. So why is option B 21.2? Wait, maybe the diagram is for one cable, and the problem statement's "two cables" is a mistake. Alternatively, maybe the angle is with the vertical? No, it says "with the horizontal". Wait, let's check the calculation again. \( \sin(45°) \approx 0.7071 \), \( 30 \times 0.7071 \approx 21.21 \), which is option B. So maybe the problem has one cable, not two? Maybe a misstatement. So the weight is equal to the total vertical tension. If one cable, \( W = 30\sin(45°) \approx 21.2\,\text{N} \) (option B). But the problem says two cables. Wait, maybe the diagram is two cables, each with 30N, angle 45° to horizontal, so vertical components add. But 2×21.2=42.4, which is option D. The options include D:42.4. So which is correct?

Wait, let's re-express:

For equilibrium, the sum of vertical forces is zero. The weight \( W \) acts downward, and the vertical components of the cables' tensions act upward.

If there are two cables, each with tension \( T = 30\,\text{N} \), angle \( \theta = 45^\circ \) to horizontal, then vertical component of each tension is \( T\sin\theta \). So total upward force: \( 2T\sin\theta \). This must equal \( W \) (downward).

So \( W = 2 \times 30 \times \sin(45^\circ) = 60 \times \frac{\sqrt{2}}{2} = 30\sqrt{2} \approx 42.4\,\text{N} \), which is option D.

But the diagram shows "30N" and "45°", maybe a single cable? Then \( W = 30\sin(45^\circ) \approx 21.2\,\text{N} \), option B.

The problem says "two cables", so likely D. But let's check the options. The options are A:10.2, B:21.2, C:32.4, D:42.4.

Wait, maybe the diagram is a single cable, and the problem's "two cables" is an error. Let's see: if it's one cable, \( W = 30\sin(45°) \approx 21.2\), which is option B. If two cables, \( 2 \times 21.2 = 42.4 \), option D.

Given that the diagram shows one cable (with 30N, 45°), maybe the problem meant one cable. So the weight is the vertical component of the tension: \( W = T\sin\theta = 30 \times \sin(45^\circ) \approx 30 \times 0.7071 \approx 21.2\,\text{N} \), which is option B. Wait, but the problem says two cables. This is confusing. Alternatively, maybe the angle is with the vertical? If angle with vertical is 45°, then horizontal is 45° from vertical, so angle with horizontal is 45°, same as before. Wait, no.

Wait, let's calculate \( 30\sin(45°) \approx 21.2 \), which is option B. So maybe the answer is B.

Answer:

B. 21.2N