Question

= 32° with the horizontal. A block of weight W = 280 N can be moved anywhere along the bar; its position is defined by the distance x = 1.10 m from In the figure, a thin horizontal bar AB of negligible weight and length L = 3.6 m is hinged to a vertical wall at A and supported at B by a thin wire BC that makes an angle the wall to its center of mass. Find (a) the tension in the wire, and the (b) horizontal and (c) vertical components of the force on the bar from the hinge at A. COM O SDB La Number Units

Answer 1

Sol" :- Given. $ = 32 l = 3.6 m xelilom Hinged at A weight of block, W=280N RASA 4 I sino N TGor' @ Tating Force and moment equilibrium, Ray +T. sing = 280 (1) Now, moment about 'A, EMATO la » T Sin 32°)* (316) = 280 X 1.10 oi T = 280x1:10 to 3.6 x Sin(3201 Det d a T=161145 ) . This is tension in wire. : 1- 3.6 x Sin(2]

Opomponent of b) At A The horizontal reaction force Rax = T Cose _ =161045 x wf32° I Rad = 136.92 N - ② vertial component of force at A ie Ray From lij Ray = 280-T sino = 280 – 167.45 X Sin 32° I Ray = 199.44 m I