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Grade: 11

                        

A 74.6-kg window cleaner uses a 10.3-kg ladder that is 5.12 m long. He places one end 2.45 m from a wall and rests the upper end against a cracked window and climbs the ladder. He climbs 3.10 m up the ladder when the window breaks. Neglecting friction between the ladder and the window and assuming that the base of the ladder does not slip, find (a) the force exerted on the window by the ladder just before the window breaks and (b) the magnitude and direction of the force exerted on the ladder by the ground just before the window breaks.

5 years ago

Answers : (1)

Kevin Nash
askIITians Faculty
332 Points
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Round off to three significant figures,
Fw = 269 N
Therefore, the force exerted by the ladder on the window is 269 N .


234-2155_1.PNG

Therefore, the vertical force exerted by the ground on the ladder is 832 N.
The horizontal force from the ground is equal to the force Fw, therefore
Fx = 269 N

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234-1207_1.PNG
5 years ago
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  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


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