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Question

# A uniform ladder of mass 10kg leans against a smooth vertical wall making an angle of 53 with it. The other end rests on a rough horizontal floor. Find the normal force and the frictional force that the floor exerts on the ladder

A
65N,65N,98N
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B
65N,98N,65N
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C
98N,65N,65N
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D
65N,65N,65N
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Solution

## The correct option is B 65N,98N,65NGiven :- Mass of ladder m = 10 kg Let, Acceleration due to gravity (g) = 9.8 m/s2 Angle between ladder and wall = 53°To Find :- 1. Normal forces (N1andN2) 2. Frictional force (f)Solution :- since, the ladder is in balanced condition ∴ N1=f (N1⇆f) And, N2=W=mg (N2↑↓W) Now, finding Torque on ladder about A (53°) we get, N1(AB)cos53°=W(AB2)sin53° N1×35=mg2×45 ∴ N1=98×23⟹65N––––– Now, N2=mg⟹98N––––– And, N1=f⟹65N––––– Hence , Option B(65N,98N,65N) is correct.–––––––––––––––––––––––––––––––––––––––––

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