Give step-by-step solution with explanation and final answer:To increase the volume of a stereo, a person applies the forces Na, £4, Ng, and Fgto the volume control knob. All
of these forces lie in t

Question

Give step-by-step solution with explanation and final answer:To increase the volume of a stereo, a person applies the forces Na, £4, Ng, and Fgto the volume control knob. All
of these forces lie in the same plane, which is parallel to the yz plane, and N4 and Ng have the same line of action,
which passes through the x axis, £4 is perpendicular to N, and Fgis perpendicular to Ng. To prevent the knob
from slipping, the person applies the forces so that £4 = 0.2N4 and Fg= 0.2Ng. Take the value of Nato be 5 N.
NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part.
5
12mm
25°
“il
Ax
AVE
Na%r 20, x
pA
If Na and the forces applied by the person are to be a couple or a system of couples, determine the values of £4, Ng, and Fg.
The value of F4is[___]N. (Round the final answer to two decimal places.)
The value of Ngis[____|N. (Round the final answer to one decimal place.)
The value of Fgis[ |. (Round the final answer to two decimal places.)

VIPSolutions · Accepted Answer

Let's break down the problem step by step:

---

## **Given Data**

- $ N_A $ = **5 N**
- $ F_A = .2 N_A $
- $ F_B = .2 N_B $
- All forces are in the same plane (parallel to the yz plane).
- $ N_A $ and $ N_B $ have the same line of action at an angle of 25°.
- $ F_A $ is perpendicular to $ N_A $, and $ F_B $ is perpendicular to $ N_B $.
- Distances: $ OA = 20 $ mm, $ OB = 12 $ mm.

---

## **Step 1: Calculate $ F_A $**

$$
F_A = .2 N_A = .2 \times 5 = 1.00~\text{N}
$$

---

## **Step 2: Analyze Couples**

For a system of couples, the net force must be zero, and the sum of the moments (couples) must be zero.

### **Force Balance (Net Force = ):**

Since all forces are in the same plane and arranged as described, the forces should balance:

$$
N_A + F_A + N_B + F_B = 
$$

But since each $ N $ and $ F $ act in orthogonal pairs (couples), we'll focus on moment arms.

---

## **Step 3: Moment About the Center (O)**

Let’s use the perpendicular distances from O to the lines of action:

- For $ N_A, F_A $ at A: distance $ r_A = 20 $ mm = .02 m
- For $ N_B, F_B $ at B: distance $ r_B = 12 $ mm = .012 m

Each pair forms a couple:
- Magnitude of couple at A: $ C_A = N_A \cdot d_{F_A} = N_A \times r_A $
- Magnitude of couple at B: $ C_B = N_B \cdot d_{F_B} = N_B \times r_B $

For the system to be a couple only, the total moment must be zero:

$$
N_A \cdot .02 = N_B \cdot .012
$$

$$
N_B = \frac{N_A \cdot .02}{.012} = \frac{5 \times .02}{.012} = \frac{.1}{.012} = 8.33~\text{N}
$$

---

## **Step 4: Calculate $ F_B $**

$$
F_B = .2 N_B = .2 \times 8.33 = 1.67~\text{N}
$$

---

## **Step 5: **Final Answers**

| Variable | Value           | Rounded Value    |
|----------|-----------------|-----------------|
| $ F_A $ | 1.00 N          | 1.00 N          |
| $ N_B $ | 8.33 N          | 8.3 N           |
| $ F_B $ | 1.666... N      | 1.67 N          |

---

## **Summary Table**

| Variable | Final Answer      |
|----------|------------------|
| $ F_A $ | **1.00 N**        |
| $ N_B $ | **8.3 N**         |
| $ F_B $ | **1.67 N**        |

---

### **Boxed Answers**

- The value of $ F_A $ is **1.00 N**
- The value of $ N_B $ is **8.3 N**
- The value of $ F_B $ is **1.67 N**

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