A cart weighing 60 lb is placed on a ramp inclined at 15° to the horizontal. The cart is held in place by a rope inclined at 60° to the horizontal, as shown in the figure. Find the force that the rope must exert on the cart to keep it from rolling down the ramp. (Round your answer to one decimal place in lbs.)

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

the correct answer is d

hope this

lockers 1, 4, 9, 16, 25, 40, and 49 are open

i ran through it by the other answer is wrong