**Problem Statement:**A uniform ladder 10.0 m long is leaning against a frictionless wall at an angle of \(60.0^\circ\) above the horizontal. The weight of the ladder is 50.0 N. A 150.0 N boy climbs 4.00 m up the ladder. What is the magnitude of the frictional force exerted on the ladder by the floor?**Solution:**1. **Determine Forces Acting on the Ladder:** - The weight of the ladder, \( W_L = 50.0 \text{ N} \) - The weight of the boy, \( W_B = 150.0 \text{ N} \) - The perpendicular distance to the ladder's center of gravity from the wall (half the ladder's length due to uniformity) is \( \frac{10.0 \, \text{m}}{2} \cos(60^\circ) = 5.0 \, \text{m} \cos(60^\circ) \). - The perpendicular distance to the boy is \( 4.0 \, \text{m} \cos(60^\circ) \).2. **Assume Equilibrium Conditions:** - Sum of horizontal forces equals zero. - Sum of vertical forces equals zero. - Sum of moments about the bottom of the ladder equals zero (to find frictional force).3. **Vertical Forces:** - \( F_N - W_L - W_B = 0 \) - \( F_N = W_L + W_B \) - \( F_N = 50.0 + 150.0 = 200.0 \text{ N} \) (Normal force from the floor).4. **Horizontal Forces:** - \( F_f = F_W \) and \( F_W = F_N \tan(60^\circ) \) - \( F_f \) = frictional force from the floor - \( F_W \) = force from the wall5. **Moments About the Bottom of the Ladder:** - \( W_L \times 5.0 \cos(60^\circ) + W_B \times 4.0 \cos(60^\circ) = F_W \times 10.0 \sin(60^\circ) \) - \( 50.0 \times

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6. A uniform ladder 10.0 m long is leaning against a frictionless wall at an angle of 60.0° above the horizontal. The weight of the ladder is 50.0 N. A 150.0 N boy climbs 4.00 m up the ladder. What is the magnitude of the frictional force exerted on the ladder by the floor?

6. A uniform ladder 10.0 m long is leaning against a frictionless wall at an angle of 60.0° above the horizontal. The weight of the ladder is 50.0 N. A 150.0 N boy climbs 4.00 m up the ladder. What is the magnitude of the frictional force exerted on the ladder by the floor?

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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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Torque

Torque is the rotational equivalent of linear force in physics and mechanics. Based on the course of study, it is also known as the moment of force, rotational force, or turning effect.

Question

**Problem Statement:**

A uniform ladder 10.0 m long is leaning against a frictionless wall at an angle of \(60.0^\circ\) above the horizontal. The weight of the ladder is 50.0 N. A 150.0 N boy climbs 4.00 m up the ladder. What is the magnitude of the frictional force exerted on the ladder by the floor?

**Solution:**

1. **Determine Forces Acting on the Ladder:**
- The weight of the ladder, \( W_L = 50.0 \text{ N} \)
- The weight of the boy, \( W_B = 150.0 \text{ N} \)
- The perpendicular distance to the ladder's center of gravity from the wall (half the ladder's length due to uniformity) is \( \frac{10.0 \, \text{m}}{2} \cos(60^\circ) = 5.0 \, \text{m} \cos(60^\circ) \).
- The perpendicular distance to the boy is \( 4.0 \, \text{m} \cos(60^\circ) \).

2. **Assume Equilibrium Conditions:**
- Sum of horizontal forces equals zero.
- Sum of vertical forces equals zero.
- Sum of moments about the bottom of the ladder equals zero (to find frictional force).

3. **Vertical Forces:**
- \( F_N - W_L - W_B = 0 \)
- \( F_N = W_L + W_B \)
- \( F_N = 50.0 + 150.0 = 200.0 \text{ N} \) (Normal force from the floor).

4. **Horizontal Forces:**
- \( F_f = F_W \) and \( F_W = F_N \tan(60^\circ) \)
- \( F_f \) = frictional force from the floor
- \( F_W \) = force from the wall

5. **Moments About the Bottom of the Ladder:**
- \( W_L \times 5.0 \cos(60^\circ) + W_B \times 4.0 \cos(60^\circ) = F_W \times 10.0 \sin(60^\circ) \)
- \( 50.0 \times

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