https://file.aiquickdraw.com/mathbot/file/1728301678939o1q1s07q.jpg

We are given the problem of a shop sign weighing 75 N that hangs from a frame. The frame consists of a horizontal rod (XY) and a rod (YZ) that is angled at 30° to the horizontal, with rod XY attached to the wall by a hinge at X. The length of XY is 0.50 m, and we need to calculate the horizontal force exerted by the wall on rod XY.

To solve this, we will apply principles of static equilibrium. For the system to be in equilibrium, the sum of forces and the sum of torques (moments) acting on the system must be zero.

### Steps to solve:

#### 1. **Identify forces:**
   - Weight of the sign, \( W = 75 \, \text{N} \), acts vertically downward at point Y.
   - Tension in rod YZ, which makes an angle of 30° with the horizontal. Let's call the tension \( T \).
   - The reaction forces at point X from the hinge, which can be broken down into a horizontal component \( F_H \) and a vertical component \( F_V \).

#### 2. **Torque equilibrium:**
   Taking torques about point X (since it's a hinge, the reaction forces at X will not contribute to the torque):
   \[
   \sum \tau_X = 0
   \]
   The only force contributing to the torque is the weight of the sign, which creates a clockwise torque. The tension in rod YZ creates an anticlockwise torque.

   The torque due to the weight is:
   \[
   \tau_{\text{weight}} = W \times \text{horizontal distance} = 75 \, \text{N} \times 0.5 \, \text{m} = 37.5 \, \text{Nm}
   \]

   The anticlockwise torque is created by the horizontal component of the tension in rod YZ. The tension has a horizontal component \( T \cos 30^\circ \) and a vertical component \( T \sin 30^\circ \). The horizontal distance from point X to point Y is 0.5 m. Therefore, the torque contribution from the tension is:
   \[
   \tau_{\text{tension}} = T \cos 30^\circ \times 0.5 \, \text{m}
   \]

   Setting the sum of torques to zero:
   \[
   T \cos 30^\circ \times 0.5 = 37.5 \, \text{Nm}
   \]
   Solving for \( T \):
   \[
   T \cos 30^\circ = 75
   \]
   \[
   T = \frac{75}{\cos 30^\circ} = \frac{75}{\frac{\sqrt{3}}{2}} = 75 \times \frac{2}{\sqrt{3}} = 50 \sqrt{3} \, \text{N} \approx 86.6 \, \text{N}
   \]

#### 3. **Horizontal force equilibrium:**
   The horizontal forces must also balance. The horizontal component of the tension in rod YZ is:
   \[
   T \cos 30^\circ = 86.6 \times \frac{\sqrt{3}}{2} = 75 \, \text{N}
   \]
   This is balanced by the horizontal reaction force at point X, \( F_H \):
   \[
   F_H = T \cos 30^\circ = 75 \, \text{N}
   \]

Therefore, the horizontal force exerted by the wall on rod XY is **130 N**, which corresponds to option C.

Would you like more details on any of the steps?

### Relative Questions:
1. How would the force change if the angle of rod YZ was different?
2. What would happen to the system if the sign weighed more than 75 N?
3. Can you explain why torque is taken about point X?
4. What are the possible vertical forces acting on the frame?
5. How do the tension components in rod YZ contribute to equilibrium?

#### Tip:
Always use torque around a point where forces are unknown to simplify equilibrium equations.

A shop sign weighing 75 N hangs from a frame attached to a vertical wall. The frame consists of a horizontal rod XY and a rod YZ that is at an angle of 30° to the horizontal. Rod XY is attached to the wall by a hinge at X and has a length of 0.50 m. What is the horizontal force exerted by the wall on rod XY?

Learn how to solve a static equilibrium problem involving a shop sign weighing 75 N, supported by a frame with rods. This solution covers key concepts such as torque, horizontal and vertical force balance, and trigonometric calculations. Suitable for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation