Question 1113342
50 lb=22.68 kg


The angle between the rope and the ramp is 60°-15°=45° 


For the cart to be in equilibrium, the component of the weight directed parallel to the incline and pointing downward must equal the component of the tension parallel to the incline and pointing upwards


T cos 45°  = mg sin Θ


T= mg sin Θ/cos 45° = (22.68*9.8*sin 15)/cos 45°  = 81.4 N (Answer) 


My solution is definitely correct. Ignore the solution by Ikleyn, as it is pure nonsense. There is no normal component of the rope's tension.