VIPSolutionsChapter 12: Homework Problems Sond Help Save & Exit 16 Required information Part1or2 Problem 12.018 - Blocks connected by a pulley system Block A has a mass of 70 kg, and block 8 has a mass of 14 kg. The coefficients of friction between all surfaces of contact are pt, = 020 and py = 0.15 294 points eBook Hit & pr p References Problem 12.018.a - Acceleration of the block in the pulley system If P= 0, determine the acceleration of block B. (Round the final answer to three decimal places.) The acceleration of block Bis J mis” 1 25°.
Question:
Chapter 12: Homework Problems Sond Help Save & Exit
16
Required information
Part1or2 Problem 12.018 - Blocks connected by a pulley system
Block A has a mass of 70 kg, and block 8 has a mass of 14 kg. The coefficients of friction between all surfaces of contact
are pt, = 020 and py = 0.15
294
points
eBook
Hit
&
pr p
References
Problem 12.018.a - Acceleration of the block in the pulley system
If P= 0, determine the acceleration of block B. (Round the final answer to three decimal places.)
The acceleration of block Bis J mis” 1 25°.
This is AI generated Response.
Asked by: Satish Reddy
Created at: 2025-06-27 05:02:22
Asked by: Satish Reddy
Created at: 2025-06-27 05:02:22
Final Answer:
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Let's break down the problem:
### **Given:**
- Mass of Block A (\(m_A\)) = 70 kg
- Mass of Block B (\(m_B\)) = 14 kg
- Coefficient of static friction (\(\mu_s\)) = 0.20
- Coefficient of kinetic friction (\(\mu_k\)) = 0.15
- Incline angle (\(\theta\)) = 25°
- \(P = 0\) (no applied force)
### **Required:**
Find the acceleration of block B (\(a_B\)).
---
## **Step 1: Forces on Block A (along the incline)**
Block A is on an incline:
- Gravity component down the incline: \(m_A g \sin\theta\)
- Normal force: \(N = m_A g \cos\theta\)
- Friction force: \(F_{friction} = \mu_k N\)
- Tension in the rope: \(T\) (up the incline)
Eq. of motion for Block A (up the incline as positive):
\[
m_A a = T - m_A g \sin\theta - F_{friction}
\]
But friction opposes motion:
\[
F_{friction} = \mu_k m_A g \cos\theta
\]
So,
\[
m_A a = T - m_A g \sin\theta - \mu_k m_A g \cos\theta
\]
---
## **Step 2: Forces on Block B (vertical)**
Block B is hanging:
- Gravity: \(m_B g\)
- Tension: \(T\)
Eq. of motion for Block B (down is positive):
\[
m_B a = m_B g - T
\]
---
## **Step 3: Combine Equations**
Add the two equations (eliminate \(T\)):
\[
m_A a + m_B a = m_B g - [m_A g \sin\theta + \mu_k m_A g \cos\theta]
\]
\[
(m_A + m_B)a = m_B g - m_A g \sin\theta - \mu_k m_A g \cos\theta
\]
\[
a = \frac{m_B g - m_A g \sin\theta - \mu_k m_A g \cos\theta}{m_A + m_B}
\]
---
## **Step 4: Plug in Values**
- \(m_A = 70 \, \text{kg}\)
- \(m_B = 14 \, \text{kg}\)
- \(\mu_k = 0.15\)
- \(g = 9.81 \, \text{m/s}^2\)
- \(\sin 25^\circ \approx 0.4226\)
- \(\cos 25^\circ \approx 0.9063\)
Plug in:
\[
a = \frac{14 \times 9.81 - 70 \times 9.81 \times 0.4226 - 0.15 \times 70 \times 9.81 \times 0.9063}{70 + 14}
\]
Calculate each term:
- \(14 \times 9.81 = 137.34\)
- \(70 \times 9.81 \times 0.4226 \approx 289.97\)
- \(0.15 \times 70 \times 9.81 \times 0.9063 \approx 93.47\)
- \(70 + 14 = 84\)
Now combine:
\[
a = \frac{137.34 - 289.97 - 93.47}{84}
\]
\[
a = \frac{137.34 - 383.44}{84}
\]
\[
a = \frac{-246.10}{84}
\]
\[
a \approx -2.93 \, \text{m/s}^2
\]
---
### **Final Answer:**
\[
\boxed{-2.93 \ \text{m/s}^2}
\]
(The negative sign indicates the direction opposite the assumed positive, so block B accelerates **upward** relative to the direction we chose as positive for B.)
Question:
generate diagram
This is AI generated Response.
Asked by: Satish Reddy
Created at: 2025-06-27 05:03:30
Asked by: Satish Reddy
Created at: 2025-06-27 05:03:30
Final Answer:
Rendered
Raw Text
Did you like this answer?
Certainly! Here's a diagram illustrating the pulley system with the two blocks:
```
Inclined Plane
/|
/ |
/ |
/ |
A/____|____
/ \
/ \
Normal Tension T
Force |
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