SOLUTION: 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 t
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Question 1198182: 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 Answer by ikleyn(52815) (Show Source):
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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.
