Let's solve the problem step by step: **Given:** - A uniform ladder (slender rod), weight \( W = 30 \) lb, length \( L = 10 \) ft. - Released from rest in the vertical position. - Friction at point A is neglected. - Find the angle \( \theta \) at which the bottom end \( A \) starts to slide. --- ## 1. **Free Body Diagram and Forces** Since friction at \( A \) is neglected, the only forces acting are: - Weight (\( W \)), acting at the center of gravity (middle of the ladder). - Normal reaction at \( A \) (vertical). - Normal reaction at the wall (horizontal, if present). But since the wall isn't holding the ladder (ladder is just falling), only gravity acts on the center of mass. --- ## 2. **Kinematics and Dynamics** We need to find the angle \( \theta \) at which the normal reaction at \( A \) (vertical) becomes zero, i.e., when the base just loses contact with the ground. For a slender rod rotating about point \( A \): - Moment of inertia about \( A \): \[ I_A = \frac{1}{3} m L^2 \] where \( m \) is the mass. - Center of mass is at \( L/2 \) from \( A \). --- ## 3. **Equations of Motion** Let \( \theta \) be the angle the ladder makes with the ground. ### a. **Force Analysis at Point A** The vertical reaction at \( A \) is \( N_A \). The ladder will lose contact when \( N_A = 0 \). ### b. **Tangential Acceleration of Center of Mass** The motion is rotational about point \( A \): \[ \sum M_A = I_A \alpha = (W) \left(\frac{L}{2}\right) \sin\theta \] \[ I_A \alpha = m g \frac{L}{2} \sin\theta \] \[ \frac{1}{3} m L^2 \alpha = m g \frac{L}{2} \sin\theta \] \[ \alpha = \frac{3g}{2L} \sin\theta \] ### c. **Translational Acceleration of Center of Mass** The acceleration of the center of mass (G) has radial and tangential components. \[ a_G = a_{G_t} \hat{t} + a_{G_n} \hat{n} \] Where: \[ a_{G_t} = \frac{L}{2} \alpha \] \[ a_{G_n} = \frac{L}{2} \omega^2 \] ### d. **Vertical Forces at A** \[ \sum F_y = m a_{G_y} \] Where \( a_{G_y} \) is the vertical component of the acceleration of the center of mass. \[ N_A + 0 = m a_{G_y} + m g \] But since \( N_A \) is upward and gravity is downward: \[ N_A = m (a_{G_y} + g) \] When the ladder loses contact, \( N_A = 0 \): \[ a_{G_y} = -g \] --- ## 4. **Finding \( a_{G_y} \) in Terms of \( \theta \)** From kinematics, the center of mass moves on a circular arc about \( A \): \[ a_{G_y} = a_{G_t} \cos\theta - a_{G_n} \sin\theta \] \[ a_{G_y} = \left( \frac{L}{2} \alpha \right) \cos\theta - \left( \frac{L}{2} \omega^2 \right) \sin\theta \] --- ## 5. **Energy Conservation to Find \( \omega^2 \)** Use conservation of energy from vertical (θ = 0) to arbitrary angle θ: \[ \Delta KE = -\Delta PE \] Initial PE (vertical): \( m g (L/2) \) At angle θ: \( m g (L/2) \sin\theta \) \[ \frac{1}{2} I_A \omega^2 = m g \frac{L}{2} (1 - \sin\theta) \] \[ \frac{1}{2} \left(\frac{1}{3} m L^2\right) \omega^2 = m g \frac{L}{2} (1 - \sin\theta) \] \[ \frac{1}{6} m L^2 \omega^2 = m g \frac{L}{2} (1 - \sin\theta) \] \[ \omega^2 = \frac{3g}{L}(1 - \sin\theta) \] --- ## 6. **Substitute and Solve for \( \theta \):** Recall: \[ a_{G_y} = \frac{L}{2} \alpha \cos\theta - \frac{L}{2} \omega^2 \sin\theta \] Plug in \( \alpha = \frac{3g}{2L} \sin\theta \) and \( \omega^2 = \frac{3g}{L}(1-\sin\theta) \): \[ a_{G_y} = \frac{L}{2} \left(\frac{3g}{2L} \sin\theta\right) \cos\theta - \frac{L}{2} \left(\frac{3g}{L}(1-\sin\theta)\right) \sin\theta \] \[ a_{G_y} = \frac{3g}{4} \sin\theta \cos\theta - \frac{3g}{2} \sin\theta (1 - \sin\theta) \] \[ a_{G_y} = \frac{3g}{4} \sin\theta \cos\theta - \frac{3g}{2} \sin\theta + \frac{3g}{2} \sin^2\theta \] Set \( a_{G_y} = -g \): \[ \frac{3g}{4} \sin\theta \cos\theta - \frac{3g}{2} \sin\theta + \frac{3g}{2} \sin^2\theta = -g \] Divide both sides by \( g \): \[ \frac{3}{4} \sin\theta \cos\theta - \frac{3}{2} \sin\theta + \frac{3}{2} \sin^2\theta + 1 = 0 \] --- ## 7. **Final Equation to Solve for \( \theta \):** \[ \frac{3}{4} \sin\theta \cos\theta - \frac{3}{2} \sin\theta + \frac{3}{2} \sin^2\theta + 1 = 0 \] Let’s use substitution \( \sin\theta = x \), \( \cos\theta = \sqrt{1-x^2} \): \[ \frac{3}{4} x \sqrt{1-x^2} - \frac{3}{2} x + \frac{3}{2} x^2 + 1 = 0 \] This is a transcendental equation and is best solved numerically. --- ## 8. **Numerical Solution:** Let’s estimate \( \theta \): Try \( \theta = 60^\circ \) (\( \sin 60^\circ = 0.866 \), \( \cos 60^\circ = 0.5 \)): \[ \frac{3}{4}(0.866)(0.5) - \frac{3}{2}(0.866) + \frac{3}{2}(0.866^2) + 1 \] \[ \frac{3}{4}(0.433) - 1.299 + 1.125 + 1 \] \[ 0.32475 - 1.299 + 1.125 + 1 = 1.14975 \] Try \( \theta = 45^\circ \) (\( \sin 45^\circ = 0.707 \), \( \cos 45^\circ = 0.707 \)): \[ \frac{3}{4}(0.707)(0.707) - \frac{3}{2}(0.707) + \frac{3}{2}(0.5) + 1 \] \[ \frac{3}{4}(0.5) - 1.0605 + 0.75 + 1 \] \[ 0.375 - 1.0605 + 0.75 + 1 = 0.0645 \] Try \( \theta = 40^\circ \) (\( \sin 40^\circ \approx 0.643 \), \( \cos 40^\circ \approx 0.766 \)): \[ \frac{3}{4}(0.643)(0.766) - \frac{3}{2}(0.643) + \frac{3}{2}(0.413) + 1 \] \[ \frac{3}{4}(0.492) - 0.9645 + 0.6195 + 1 \] \[ 0.369 - 0.9645 + 0.6195 + 1 = 1.024 \] Try \( \theta = 30^\circ \) (\( \sin 30^\circ = 0.5 \), \( \cos 30^\circ = 0.866 \)): \[ \frac{3}{4}(0.5)(0.866) - \frac{3}{2}(0.5) + \frac{3}{2}(0.25) + 1 \] \[ \frac{3}{4}(0.433) - 0.75 + 0.375 + 1 \] \[ 0.32475 - 0.75 + 0.375 + 1 = 0.94975 \] Try \( \theta = 75^\circ \) (\( \sin 75^\circ = 0.9659 \), \( \cos 75^\circ = 0.2588 \)): \[ \frac{3}{4}(0.9659)(0.2588) - \frac{3}{2}(0.9659) + \frac{3}{2}(0.933) + 1 \] \[ \frac{3}{4}(0.250) - 1.449 + 1.3995 + 1 \] \[ 0.1875 - 1.449 + 1.3995 + 1 = 1.138 \] Try \( \theta = 80^\circ \) (\( \sin 80^\circ \approx 0.9848 \), \( \cos 80^\circ \approx 0.1736 \)): \[ \frac{3}{4}(0.9848)(0.1736) - \frac{3}{2}(0.9848) + \frac{3}{2}(0.9698) + 1 \] \[ \frac{3}{4}(0.171) - 1.4772 + 1.4547 + 1 \] \[ 0.12825 - 1.4772 + 1.4547 + 1 = 1.10575 \] Try \( \theta = 90^\circ \) (\( \sin 90^\circ = 1 \), \( \cos 90^\circ = 0 \)): \[ \frac{3}{4}(1)(0) - \frac{3}{2}(1) + \frac{3}{2}(1) + 1 = 0 - 1.5 + 1.5 + 1 = 1 \] Try lower values. Let's try \( \theta = 10^\circ \) (\( \sin 10^\circ \approx 0.174 \), \( \cos 10^\circ \approx 0.985 \)): \[ \frac{3}{4}(0.174)(0.985) - \frac{3}{2}(0.174) + \frac{3}{2}(0.030) + 1 \] \[ \frac{3}{4}(0.171) - 0.261 + 0.045 + 1 \] \[ 0.12825 - 0.261 + 0.045 + 1 = 0.91225 \] So, at 45°, value is close to 0. **Therefore, the angle is approximately:** \[ \boxed{45^\circ} \] --- ### **Final Answer:** **The bottom end \( A \) starts to slide when the ladder makes an angle of approximately \( \boxed{45^\circ} \) with the ground.**