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Given data The mass is m -43 The length is L 1.68 m Using the figure is .7 kg tan-y 0- tan-3.87 3.30 49.5 Equate the net torque about M to be equal to zero Tsin (BM)-g(AM)-g(BM)-0 "g(AM+BM) BM Sİn@ - 1m g tah (43.7 kg)(9.81 m/s)1,68m+3.3.m 3.3 m)sin 49.5° T-353.69 N The horizontal component of force is (353.69) cos 49.5° 229.7 N The direction of horizontal component of force will be right side because it is the normal force which acts away from the wall The vertical component of force is =mg-Tsin θ (43.7 kg) (9.81 m/s)-(353.69 N)sin 49.5° 159.75 N The direction of force will be upward

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