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A cart weighing 40 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
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There is no any figure attached to the post.
The force that the rope exerts on the cart is called "rope tension", or simply "tension".
Let F be the tension.
Then it takes off F*sin(60°) from the weight W of the cart, leaving only W-F*sin(60°) as
the net vertical force acting on the cart.
This vertical force W-F*sin(60°) can be decomposed by the usual way into the sum
of two vectors/(components), one parallel to the incline and the other normal to it.
The component directed along the incline is (W-F*sin(60°))*sin(15°).
This force/(component) is balanced by the component F*cos(60°- 15°) = F*cos(45°) directed along
the incline in the opposite direction.
Thus the equilibrium equation is
(W-F*sin(60°) )*sin(15°) = F*cos(45°).
which gives
F = .
Further, sin(15°) = 0.2588;
sin(60°)*sin(15°) + cos(45°) = + = 0.9312,
so F = = 11.12 pounds.
Answer. The tension of the rope is 11.12 pounds.