Two blocks of mass ma = 10 kg and mg = 5 kg are connected by a massless string that passes over a pulley as shown in the figure. The system is in static equilibrium. There is friction between ma and the inclined surface (m=0.4). Neglect the friction between the string and the pulley. Determine the tension in the string. 10 kg 5.0 kg 37° 98.0 N Do 52.3 N 65.35 N
Two blocks of mass ma = 10 kg and mg = 5 kg are connected by a massless string that passes over a pulley as shown in the figure. The system is in static equilibrium. There is friction between ma and the inclined surface (m=0.4). Neglect the friction between the string and the pulley. Determine the tension in the string. 10 kg 5.0 kg 37° 98.0 N Do 52.3 N 65.35 N
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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two other options could be
147.0 N
49.0 N
**Diagram Description:**
The diagram shows two blocks connected by a string passing over a pulley. Block \( m_A \) (10 kg) is on an inclined surface at an angle of \( 37^\circ \) to the horizontal. Block \( m_B \) (5 kg) hangs vertically. The friction coefficient (\( \mu_s \)) between \( m_A \) and the incline is given as 0.4.
**Answer Options:**
- \( \mathbf{98.0 \, \text{N}} \)
- \( \mathbf{0 \, \text{N}} \)
- \( \mathbf{52.3 \, \text{N}} \)
- \( \mathbf{65.35 \, \text{N}} \)
**Explanation for Educational Purposes:**
To solve for the tension in the string, we need to analyze the forces acting on both blocks and account for static equilibrium conditions.
1. **Block \( m_A \)** on the inclined plane:
- Weight (\( W_A \)) = \( m_A \cdot g \). \( W_A = 10 \cdot 9.8 = 98 \, \text{N} \).
- The component of \( W_A \) parallel to the incline: \( W_{\parallel A} = W_A \cdot \sin(37^\circ) \).
- The component of \( W_A \) perpendicular to the incline: \( W_{\perp A} = W_A \cdot \cos(37^\circ) \).
- Maximum static friction force: \( f_s = \mu_s \cdot W_{\perp A} \).
2. **Block \( m_B \)** hanging vertically:
- Weight (\( W_B \)) = \( m_B \cdot g \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa0dedf45-3cb9-4516-add5-7f5a88a5efc9%2F3776a6d6-a306-48fc-bf7f-aab6fa149b6e%2Fwd6aste_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Two blocks of mass \( m_A = 10 \, \text{kg} \) and \( m_B = 5 \, \text{kg} \) are connected by a massless string that passes over a pulley as shown in the figure. **The system is in static equilibrium.** There is friction between \( m_A \) and the inclined surface (\( \mu_s = 0.4 \)). Neglect the friction between the string and the pulley. Determine the tension in the string.
**Diagram Description:**
The diagram shows two blocks connected by a string passing over a pulley. Block \( m_A \) (10 kg) is on an inclined surface at an angle of \( 37^\circ \) to the horizontal. Block \( m_B \) (5 kg) hangs vertically. The friction coefficient (\( \mu_s \)) between \( m_A \) and the incline is given as 0.4.
**Answer Options:**
- \( \mathbf{98.0 \, \text{N}} \)
- \( \mathbf{0 \, \text{N}} \)
- \( \mathbf{52.3 \, \text{N}} \)
- \( \mathbf{65.35 \, \text{N}} \)
**Explanation for Educational Purposes:**
To solve for the tension in the string, we need to analyze the forces acting on both blocks and account for static equilibrium conditions.
1. **Block \( m_A \)** on the inclined plane:
- Weight (\( W_A \)) = \( m_A \cdot g \). \( W_A = 10 \cdot 9.8 = 98 \, \text{N} \).
- The component of \( W_A \) parallel to the incline: \( W_{\parallel A} = W_A \cdot \sin(37^\circ) \).
- The component of \( W_A \) perpendicular to the incline: \( W_{\perp A} = W_A \cdot \cos(37^\circ) \).
- Maximum static friction force: \( f_s = \mu_s \cdot W_{\perp A} \).
2. **Block \( m_B \)** hanging vertically:
- Weight (\( W_B \)) = \( m_B \cdot g \
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