222()(75kg 10kg)(9.8m/s )8.3 10 NFMmg=+=+=×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: FFF=×+×=×223222228 1083 1088 10(..N)N)N.(c) The angle φbetween the force and the horizontal is given by tan= F3/F2= 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. F1is the force of the window, horizontal because the window is frictionless. F2and F3are components of the force of the ground on the ladder. Mis the mass of the window cleaner and mis 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 θbe the angle between the ladder and the ground. We use cos/ or sin/ dLL dLθθ==−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.