Question

7

The figure shows a uniform horizontal beam (length = 8 m, mass = 20 kg) that is pivoted at the wall, with its far end supported by a cable that makes an angle of 50 degree with the horizontal. If a store's sign (mass = 30 kg) is hung at the end of beam, find all forces on the beam.

Answer 1

The Given setup is as shown here and mass of store sign m_1 = 30 kg. mass of beam m_2 = 20 kg. and the angle theta = 50 degree, l = 8m The forces acting on the beam are F_1 by the store sign: F_2 is the normal reaction by the wall on beam and F_3 is the Force by the cable rightarrow From the rotational equilibrium, of the beam about the point 'O' is (m_2 g) l/2 + F_1 middot l - (F_3 sin theta)l = 0 = > (200 times 8/2) + (300 times 8) - (F_3 sin 50)8 = 0} = > F_3 = 522.163 N rightarrow Balancing horizontal forces on Beam F_3 cos theta = F_2 } = > f_2 = 335.64 N so the forces on the beam are (i) It's own weight m_2g = 200 N (ii) weight of store sign F_1 = 300 N (iii) Force (Tension)