ch12-p009

ch12-p009 - 9 The forces on the ladder are shown in the...

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22 2 ( ) (75kg 10kg)(9.8m/s ) 8.3 10 N FM m g =+ = + = × The magnitude of the force of the ground on the ladder is given by the square root of the sum of the squares of its components: FF F = × + × = × 2 2 3 2 222 28 10 83 10 88 10 (. . N) N) N. (c) The angle φ between the force and the horizontal is given by tan = F 3 / F 2 = 830/280 = 2.94, so = 71º. The force points to the left and upward, 71º above the horizontal. We note that this force is not directed along the ladder. 9. The forces on the ladder are shown in the diagram on the right. F 1 is the force of the window, horizontal because the window is frictionless. F 2 and F 3 are components of the force of the ground on the ladder. M is the mass of the window cleaner and m is the mass of the ladder. The force of gravity on the man acts at a point 3.0 m up the ladder and the force of gravity on the ladder acts at the center of the ladder. Let θ be the angle between the ladder and the ground. We use cos / or sin / dL L dL θθ == to find
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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