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QUICK QUESTION!!** PIC INCLUDED* Suppose that a box is being towed up an inclined plane as shown in the figure. Find the force w needed to make the component of the force parallel to the inclined plane equal to 2.5 lb. Give answer in component form.

Mathematics
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We will need that figure to know exactly what is being asked.
Does anyone know how to solve it D=

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Other answers:

|dw:1352583686628:dw|
|dw:1352583791078:dw|
the component parallel to the plane is the magnitude of the force W time the cosine of the angle between the force and the plane, which is 30-15=15degrees
Wait but the angle given says 33 degrees?
sorry, good eye
|dw:1352584030549:dw|
ohokkk
but how would you get the component form..... like what exactly do i solve for? w?
yes
mhm, okay but how do you know what you have to minus 33 with 15?
we want the angle between the ramp and the force the ramp is at an angle of 15 deg to the horizontal the force is at an angle of 33 deg to the horizontal therefore the angle between the force and the ramp is 33-15deg
Oh okay cool, so how can you set it up to get W?cos 18 = 2.5lbs/w..?
start with cos(theta)=F/W where F is the component parallel to the plane we set theta=18 and F=2.5 now yuo can solve for W
cuz it's a right triangle...|dw:1352584629296:dw|
cos(theta)=adj/hyp=F/W
ohh Okay so W=2.63/
I dunno, let me find a calculator...
lol ok
yep, I get the same :)
oh okay but the answer says in the book <2.20,1.43> so how can i get there?
those must be the *components* of W, so it is W*cos(phi) W*sin(phi) where phi is the angle that W makes with the horizontal
|dw:1352585011969:dw|here my exaggerated drawing doesn't work so well :P but Wx and Wy are the x and y components respectively
what so youd get like (2.62 cos (18) , 2.62 sin (18)?
D:
18 is not the angle that W makes with the horizontal...
wait really then what would it be?O:

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