Calculate tensile force

Question

A special trough manufactured from solid steel plates is used to haul heavy equipment up a slope of 15 degrees to the horizontal in a mine. The coefficient of friction between the trough and Cocopah rails is 0,3. Fresh air is blown in the tunnel and the load is equivalent to 400Nm^2 of the projected area of the trough. The effective projected area of the trough is 100mm × 1400mm. The mass is 1,2t , the trough is banked over a distance of 25m in 35 seconds. Calculate the following:

1 Tensile force in the hauling cable to the haul of the trough

2. Work done after 5 seconds

3. Work done to cover the distance

4. Power required

Answer 1

1. R is the normal reaction of the loaded trough. F is the friction force opposing the motion of the trough. The projected area of the trough is 0.1*1.4 m^2=0.14m^2. Equivalent load=400/0.14N=2857N approx. The load is 1200kg (1.2 metric tonnes). The weight is 1200g N, where g is the acceleration of gravity (9.81 m/s/s approx). Resolving forces along the inclined plane, we have friction and gravity, F+(400/0.14+1200g)sin15º. Normal to the inclined plane we have R=(400/0.14+1200)cos15º. The coefficient of friction is 0.3 so F=0.3R=0.3(400/0.14+1200g)cos15º​. The hauling cable has to pull against friction and gravity=0.3(400/0.14+1200g)cos15º​+(400/0.14+1200g)sin15º=8025.5N tensile force.
2. Work = force times distance. Distance = 25*5/35=25/7m. Work = 8025.5*25/7=28662.5Nm=28662.5 Joules=28.6625kJ.
3. Total work done over 25m=200637.52 Joules=200.63752kJ.
4. Power=work/time=5732.5 watt or 5.732kW.