a uniform ladder of length 6m and 20kg rests against a smooth wall making an angle 60 degrees with the floor. A man weighing 50kg force is standing on the ladder 4m from its bottom. find reactions of wall and floor

Dear student

The figure shows the situation:

Where Ng is the reaction of ground, Nw is the reaction of the wall and f is the friction force. In the condition when ladder is at rest all forces are balanced so 
for horizontal forcesNw=f  for vertical forces Ng=20N+50Nso Ng=70 Nalso torques due to these forces are in equilibriumtorques in counter clock wise direction=torques in clock wise directionτwall=τladder+τman        all torques are taken about ground poin ANwsin60°·6=20×3sin30°+50×4sin30°Nw×32×6=60×12+200×12Nw=26063Nw=25.01 N

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