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January 18th, 2009, 01:17 PM   #1
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confused on a trig question

this is the original example:


I really can't wrap my brain around their explanation. It seems so illogical to me. Here's the reason why:

You have a 15 degree angle that intersects the line of gravity pulling the object straight down towards the center of the earth. Shouldn't this angle and this line representing the force of gravity create a perfectly acceptable right triangle?



And in this manner, the force of gravity pulling the 200 lbs down makes up one of the legs of the triangle.

But in their explanation, they seemingly add on an additional triangle to make the force of gravity the hypotenuse of the triangle.





what is the reasoning behind doing this? I don't understand the logic. Please explain why they used the 200lb force of gravity as the hypotenuse instead of the leg of the triangle.
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January 18th, 2009, 03:49 PM   #2
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Re: confused on a trig question

Think of the vector [downward force here] being like a stick on an angle. The value on a surface at any angle will be as a shadow cast on that surface.

The cosine times the value is the value in that new direction. If you think about it, the force down the slope must be less than the vertical force. Otherwise wedges, ramps, screws and so on lose their advantage. Since the hypotenuse is the longest side of a right triangle, you have your reason. The downward force is split into two forces, one down the plane and the other into the plane. As the slope increases, the force down the plane increases and that into the plane decreases. When flat, all the force is into the plane. When almost vertical, very little force will be into the plane.

[attachment=0:1xkfatme]Plane.jpg[/attachment:1xkfatme]
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January 18th, 2009, 05:04 PM   #3
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Re: confused on a trig question

Quote:
Originally Posted by Dave
Think of the vector [downward force here] being like a stick on an angle. The value on a surface at any angle will be as a shadow cast on that surface.
Sorry but I have no idea what you're trying to say here.

Quote:
Originally Posted by Dave
The cosine times the value is the value in that new direction. If you think about it, the force down the slope must be less than the vertical force. Otherwise wedges, ramps, screws and so on lose their advantage. Since the hypotenuse is the longest side of a right triangle, you have your reason. The downward force is split into two forces, one down the plane and the other into the plane. As the slope increases, the force down the plane increases and that into the plane decreases. When flat, all the force is into the plane. When almost vertical, very little force will be into the plane.
I pretty much understand what you're saying here, but I'm still confused.
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January 19th, 2009, 09:50 AM   #4
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Keep thinking about it. Also, why isn't the block moving (assuming it isn't)?
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January 19th, 2009, 07:39 PM   #5
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Re: confused on a trig question

well if the block isn't moving it's because the net force of all the forces acting on it is zero. I'm not good with physics (yet), but I would imagine the normal force of the plane would be cancelling the force of gravity pushing the block into the slope. Then the force of friction would cancel whatever force is pushing the block down the slope (what force is pushing the block down the slope? Gravity? I thought that only acts in a straight downward direction?)

I've thought about it, and this is what I'm thinking:

If you draw two lines that form a 75 degree angle like in this example, you have the option of where you want to put the 90 degree angle. You could create the 90 degree angle with the line that points straight downward, or you could create the 90 degree angle with the line that points 15 degrees of horizontal. The only reason I can think of as to why you would want to create the 90 degree angle with the line that is 15 degrees of horizontal in this case, is so that the net force acting to push the object down the slope is less than the weight of the object itself. Because apparently, the net force can't be greater than the weight of the object itself (although I'm pretty sure this is completely untrue, it seems to be the reasoning behind me not being able to comprehend this explanation)

I give up. Save me, skipjack : (
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January 20th, 2009, 06:41 AM   #6
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Re: confused on a trig question

Gravity pulls it straight down. That force is split into two components, one into the slope, one down the slope. That completes the right triangle in the diagram I offered. Each component is less than the gravity component, which is the hypotenuse.

Note: A vector can be split into ANY two directions, forming a parallelogram of forces. It is convenient to plsit it into those at right angles to each other ...especially in case such as this where one goes into the plane and one down the plane.
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January 23rd, 2009, 04:32 PM   #7
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Quote:
Originally Posted by floaty
if the block isn't moving, it's because the net force of all the forces acting on it is zero. I would imagine the normal force of the plane would be cancelling the force of gravity pushing the block into the slope. Then the force of friction would cancel whatever force is pushing the block down the slope (what force is pushing the block down the slope? Gravity? I thought that only acts in a straight downward direction?)
The diagram shows no reaction force from the slope, but there must be one. Hence the diagram is incomplete and may well be omitting a friction force as well. If the block isn't moving, the net force must be zero, and so (the component of) the net force acting down the slope is zero.

As Dave observed, any force may be considered to have components, since any force is equivalent to a system of forces (such as "components" you choose to help you solve the problem). The calculation provided for this problem ignores friction, so let's suppose there is no friction. In that case, the calculation is correct, but the block must be sliding down the slope. If there is friction, but the block is sliding down the slope anyway, you can't determine the net force down the slope without knowing the coefficient of friction.

Perhaps the problem would have been better worded if from the outset it had referred to the component of the gravitational force acting down the slope. So where does all this leave your alternative approach? An alternative approach is possible, but you're much more likely to do it correctly (and eventually obtain the correct answers) if you consider all the forces; if you don't, it's easy to make incorrect assumptions about what forces are balancing each other.
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