Question 1198182
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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.



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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

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(W-F*sin(60°) )*sin(15°) = F*cos(45°).


which gives


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;F = {{{(W*sin(15^o))/(sin(60^o)*sin(15^o) + cos(45^o))}}}.


Further,  sin(15°) = 0.2588;


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;sin(60°)*sin(15°) + cos(45°) = {{{(sqrt(3)/2)0.2588}}} + {{{sqrt(2)/2)}}} = 0.9312,


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;so  F = {{{(40*0.2588)/0.9312}}} = 11.12 pounds.


<U>Answer</U>.  The tension of the rope is 11.12 pounds.
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