A 57 kg window cleaner uses a 16.56 m uniform ladder of mass 6.2kg . He places o

Question

A 57 kg window cleaner uses a 16.56 m uniform ladder of mass 6.2kg . He places one end on the ground 2.07m from a wall and the other against a cracked window. The window breaks when the cleaner is 1.52 m from the bottom end of the ladder. Neglect friction at the window and assume the ladder does not slip at its base. Find the magnitude and direction of the force on the ladder by the ground just before the window breaks . Give the force direction as an acute angle in degrees above the ground . Hint:first find the force on the window by the ladder just before it breaks.

Explanation / Answer

Given: Mass of the window cleaner = M = 57 kg Mass of the ladder = m = 6.2 kg He places one end on the ground = 2.07m (from a wall and the other against a cracked window) The window breaks when the cleaner is = 1.52 m ( from the bottom end of the ladder) magnitude and direction of the force on the ladder by the ground just before the window breaks = F = ? Apply , Rotational equlibrium ( The sum of Torques or mometns acting on this system is 0) thus, we have mathematiclly , sum of all torques acting on this      0   - 57 (9.8) (1.52) -6.2(9.8)(16.56/12) + F(2.07) = 0     -849.072 - 503.0928 + F (2.07) = 0                                       F = 653.219 N     Mass of the window cleaner = M = 57 kg Mass of the ladder = m = 6.2 kg He places one end on the ground = 2.07m (from a wall and the other against a cracked window) The window breaks when the cleaner is = 1.52 m ( from the bottom end of the ladder) magnitude and direction of the force on the ladder by the ground just before the window breaks = F = ? Apply , Rotational equlibrium ( The sum of Torques or mometns acting on this system is 0) thus, we have mathematiclly , sum of all torques acting on this      0   - 57 (9.8) (1.52) -6.2(9.8)(16.56/12) + F(2.07) = 0     -849.072 - 503.0928 + F (2.07) = 0                                       F = 653.219 N    

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