I assume you ignore friction. The cart is held in equilibrium, so the net force on the cart is zero.
There are 3 forces acting on the cart:
• weight (magnitude w, pointing down)
• normal force (mag. n, pointing perpendicular to the ramp)
• tension in the rope (mag. t, pointing 60º from the horizontal, or equivalently 60º - 15º = 45º from the parallel direction)
Split up the forces into horizontal and vertical components. We have
• horizontal:
t cos(60º) + n cos(105º) = 0
• vertical:
n sin(105º) + (-w) = 0
(the normal force has a direction of 105º from the horizontal because it's perpendicular to the ramp, so it forms an angle of 90º with the ramp, plus the 15º inclination of the ramp itself)
We're given that w = 60 lb, so
n sin(105º) = 60 lb
n ≈ 62.1 lb
Solve for t :
t cos(60º) = -n cos(105º)
t = -n cos(105º)/cos(60º)
t ≈ 32.2 lb
