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Introductory Physics Homework Help
A ladder against a wall problem
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[QUOTE="brotherbobby, post: 6185527, member: 372061"] [ATTACH type="full" width="516px" alt="244253"]244253[/ATTACH] The diagram shows the the forces and distances. (a) Using ##\Sigma \vec F = 0## for the ladder, we have the two equations ##\underline{w_L+w_M=n_G}## and ##\underline{f_G = n_W}##. Taking moments about the ground ##G##, we have ##w_L d/2+W_M l_M \cos \theta = n_W h## which leads to ##\mathbf{n_W} = \frac{w_L d/2+w_M l_M \cos\theta}{h} = \frac{16 \times 9.8 \times 1+70\times 9.8\times 3.5\times \cos 69.1^{\circ}}{5.6 \sin 69.1^{\circ}} = \mathbf{193.69 \text{N}}## . From the underlined equations above, we get ##\mathbf{f_G = 193.69 \text{N}}## and ##\mathbf{n_G} = (16+70)\times 9.8 = \mathbf{842.8 \text{N}}##. (b) From the diagram above (see inlet), we find that the force of the ground on the ladder ##\mathbf{n} = \sqrt{n_G^2+f_G^2} = \sqrt{842.8^2+193.69^2} = \mathbf{864.77 \text{N}}##. (c) The angle that force from the ground makes with the horizontal : ##\boldsymbol{\alpha} = \tan^{-1} \frac{n_G}{f_G} = \mathbf{77.06^{\circ}}##.Reflection : 1. Argue why ##\alpha > \theta##. One way to do that is to show that if the two angles were the same then the ladder will not be in equilibrium. 2. Let the friction due to the window (wall) be some ##f_W##. Can you solve for the (four) forces? [/QUOTE]
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A ladder against a wall problem
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