Question 1113342
We presume there is no  friction then the parallel component  acting along the incline is mg sin ( 15) 
The rope is at an angle of ( 60- 15 ) = 45 degrees from the incline. 
so the component of the force of the rope along the incline is F sin(45) 

Since the cart is stationary there is equilibrium.

F sin( 45) = m * g * sin(15) 

F = 50 * sin(15) / sin(45 ) lb wt 

= 50 * 32 * sin(15) / sin ( 45) poundals.

sin 45 = 1/sqrt(2)
sin 15 deg =0.26

plug the values