6) [8 pts] In the diagram shown, the uniform rod is free to pivot about its base in the plane of the paper. It is motionless and stable at an angle of 30 degrees from the horizonta|.The rod has a mass of 4.0 kg and a length of 60.0 cm. The 3.0 kg mass is suspended 40.0 cm from the pivot end of the rod. The wire is anchored 20 cm (vertically) above the pivot point, and is horizontal. Find the tension FT in the wire. Hint: consider the different torques about the pivgt point. The moment of inertia for a rod of mass M and length L, about its end, is 1l3 ML . pivot point 3.0 kg To slove this we need to balance the tongues acting on the nod . 0.3 using the pivot in this figure; FT centre of mass = 0.3 the tongues are a the 3kq 0.3 component of the force 20 em 39 WR acts from 30 ' Perpendicular to the nod times the y pivot distance from the pivot. we will take counter clock wie Torque as Positive and clockwise as negative . The perpendicular component of the hanging mass is = mg sin 60 : . Torque at the pivot due to the mass (3kg ) - (block = - (my sin 60 ') x 0.4 = - ( 3 x 9. 8 x Sin (60)) XO. 4 = - 10.2 NM The weight of the rod ( WR ) = 4x 8 = 4x9-8 N The perpendicular Component of WR = 4x9.8 x Sin (60' ) . Torque at the pivot due to the weight of the T Rod = - 4x 9. 8 x Sin ( 60' ) x 0.3 we take 0.3 as the weight Wp acts across centre of mass = 0.3Rod = - 10.2 N.M. The distance about which the tension force acts = 20 cm yo em = 04 m Sin (30 ) The torque due to the Tension ( FT ) = ) Tension = + F- XO . 4 x Sin (30' ) = 0.2 + T since, the rod is Static , the torques must balance Tension + block f "Rod = 0 ) 0.2 FT - 10.2 - 10.2 20 FT = 20. 4 = 102 N 0 . 2 . . The tension force ( FT ) = 102 N