text: 2nd request, any ideas/help? a 588 n uniform boom (l = 24 m) is supported by a cable and is at an angle of 48 deg above the horizontal. the boom is pivoted at the bottom and a 1770 n object hangs from its end.  a support cable is attached (at a 90deg angle) to the boom at a distance of 0.66 l from the foot of the boom and its tension is perpendicular to the boom.   a) find the tension t in the support cable in units n.  b) find the horizontal component of the reaction force on the boom by the floor in units n.  c) find the vertical components of the reaction force on the boom by the floor in units n. the beam is at 48 degrees to the horizontal.. its weight of 588n acts at its middle (0.5l) vertically downwards. there is a load of 1770n at its end (1l). the tension due to the cable acts perpendicular to the beam and at a distance of 0.66l. now, the basic equation in physics tells that the clockwise moment is balanced by the counter-clockwise moment in a stationary object. so, if 't' is the tension in the cable;  t*0.66l = 588*0.5l*cos48 + 1770*1l*cos48  solve this to get the answer to a).  the answer to b) is got by equating the horizontal forces to zero as the beam is stationary. the equation is:  hr - t*cos(90-48) = 0.  here hr is the horizontal component of the reaction force by the floor. t is as above  c) the vertical component is got by equating the vertical forces to zero.  vr - 588 - 1770 + t*sin(90-48) = 0