After solving the problem also determine the normal and frictional forces at the left end of the beam. **Problem 52**An 80 kg construction worker sits down 2.0 m from the end of a 1450 kg steel beam to eat his lunch, as shown in Figure P8.52. The cable supporting the beam is rated at 15,000 N. Should the worker be worried?**Figure Explanation:**Figure P8.52 illustrates a construction setup where a steel beam is horizontally fixed to a wall on one end. A cable runs from the wall to the far end of the beam at a 30° angle with the beam. The length of the beam is 6.0 meters. A construction worker is seated 2.0 meters from the wall along the beam. The situation involves analyzing the forces acting on the beam, particularly the tension in the cable relative to its rated capacity.

Given Values Weight of the worker is 80 kg Distance of the worker from the end of the beam is 2 m Length of the beam is 6 m Weight of the beam is 1450 kg Maximum Tension that the cable can handle is 15000 N The vertical component of the distance of the beam can be calculated as Dv=6sin30Dv= 3 m Let the tension in the cable be T Now for calculating the tension in the cable we need to balance the torque The torque in the beam can be calculated as τbeam =1450×9.81×3τbeam =42673.5 Nm The torque due the worker can be calculated as τworker =(80×9.81×4)τbeam =3139.2 Nm Now the net torque about the point of contact of the wall can be calculated as τnet =τbeam+τworker3T=42673.5+3139.2T=15270.9 N Since the Tension in the cable is more than permissible tension, therefore the worker needs to worried. Now the normal force at the end can be calculated as FN=Tsin30FN=15270.9×sin30FN=7635.45 N The frictional force can be calculated as Ff=Tcos30Ff=15270.9×cos30Ff=13224.98 N