Find uso the block slides to the right at a constant speed of 2 m/s. ≤ x = 0 8 kg 30° 60 N Tx-Ff=0 Tx-ufn=0 tu Fin + tu

College Physics
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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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**Problem Statement:**
3. Find \( \mu \) so the block slides to the right at a constant speed of 2 m/s.

**Given Diagram Details:**
- An 8 kg block is on a horizontal surface.
- A force of 60 N is applied at a 30° angle above the horizontal.

**Diagram Explanation:**
- A diagram shows the forces acting on the block.
- The force vector is split into components: \( T_x \) (horizontal) and \( T_y \) (vertical).
- \( F_f \) represents the frictional force; \( F_n \) denotes the normal force.
- The weight \( mg \) acts downward.

**Equations and Calculations:**
1. **Horizontal Forces (\( \Sigma F_x = 0 \)):**
   \[
   T_x - F_f = 0 \\
   T_x = \mu F_n
   \]
   Thus, 
   \[
   T_x = \mu F_n
   \]

2. **Vertical Forces (\( \Sigma F_y = 0 \)):**
   \[
   F_n + T_y - mg = 0 \\
   F_n = mg - T_y
   \]

3. **Substituting for \( \mu \):**
   \[
   \mu = \frac{T_x}{mg - T_y}
   \]

4. **Components of Tension:**
   \[
   T_y = 60 \sin(30°) \\
   T_x = 60 \cos(30°)
   \]

This setup ensures that the block moves at constant speed, emphasizing the balance of forces in both horizontal and vertical directions.
Transcribed Image Text:**Problem Statement:** 3. Find \( \mu \) so the block slides to the right at a constant speed of 2 m/s. **Given Diagram Details:** - An 8 kg block is on a horizontal surface. - A force of 60 N is applied at a 30° angle above the horizontal. **Diagram Explanation:** - A diagram shows the forces acting on the block. - The force vector is split into components: \( T_x \) (horizontal) and \( T_y \) (vertical). - \( F_f \) represents the frictional force; \( F_n \) denotes the normal force. - The weight \( mg \) acts downward. **Equations and Calculations:** 1. **Horizontal Forces (\( \Sigma F_x = 0 \)):** \[ T_x - F_f = 0 \\ T_x = \mu F_n \] Thus, \[ T_x = \mu F_n \] 2. **Vertical Forces (\( \Sigma F_y = 0 \)):** \[ F_n + T_y - mg = 0 \\ F_n = mg - T_y \] 3. **Substituting for \( \mu \):** \[ \mu = \frac{T_x}{mg - T_y} \] 4. **Components of Tension:** \[ T_y = 60 \sin(30°) \\ T_x = 60 \cos(30°) \] This setup ensures that the block moves at constant speed, emphasizing the balance of forces in both horizontal and vertical directions.
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