Posted by **Sarah** on Sunday, March 2, 2008 at 10:15pm.

a 76 kg window cleaner uses a 12 kg ladder that is 5.8 m long. he places one end on the ground 2.8 m from a wall, rests the upper end against a cracked window, and climbs the ladder. he is 2.5 m up along the ladder when the window breaks. neglect friction between the ladder and window and assume that the base of the ladder did not slip.

A) find the magnitude of the force on the window from the ladder just before the window breaks.

B) find the magnitude and direction (angle) of the force on the ladder from the ground just before the window breaks.

Thanks a bunch!

- Physics -
**drwls**, Monday, March 3, 2008 at 1:50am
The angle that the ladder makes with the floor is cos^-1 (2.8/5.8) = 61.13 degrees

The ladder touches the window at a point 5.8 sin 61.13 = 5.079 m from the ground.

Draw a free body diagram and set the total moment of forces about the bottom of the ladder equal to zero. Three forces contribute: (1) the force of the window on ladder, (2) the weight of the window cleaner and (3), the ladder's weight. each force has a particular lever arm that you must use when calculating the moment. Moments (2) and (30 are in the opposite drection from monent (1)

(A) Solve the moment balance equation for F. The force of the ladder on the window is equal and opposite to the force of the window on the ladder.

(B) Perform a balance of vertical and horizontal forces on the ladder. At the base, the horizontal force must be equal and opposite the force F from part A. The vertical force (upward) of the floor on the ladder equals the sum of the weights of the ladder and the person on it,

(76 + 12)*9.8 = 862.4 Newtons

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