Calculus I NEED HELP!

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Robertinho!

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so the question is...

A weight of 500 pounds is supported by two cables that exert forces of -100i + 200j and 100i + 300j respectively. Find the angle between the cables.

Please explain me the process
Thanks
 
Draw a picture, I'm sure it would help you.

Do you know how the vectors you gave are drawn?
 
500 pounds is another vector, which, in this problem, you don't need to bother with.

If the lengths of the cable were given with the angles, then you would use the weight to get the vectors, and then work out the tension in each cable. But that is another story :)

Post the answers that you get!
 
Unknown008 said:
500 pounds is another vector, which, in this problem, you don't need to bother with.

If the lengths of the cable were given with the angles, then you would use the weight to get the vectors, and then work out the tension in each cable. But that is another story :)

Post the answers that you get!
Click to expand...


So we use the dot product?
 
You sure can, but if you don't want to do so, you can always use trigonometry =)

Or use trigonometry as a double check.
 
Robertinho! said:
Ok Cool! so I got 11.31 for the angle using the dot product...
Can you tell if it is right?
Click to expand...

Could you show us how you got that? I'm getting something different. And my answer is the same using dot product and trigonometry...
 
Unknown008 said:
Could you show us how you got that? I'm getting something different. And my answer is the same using dot product and trigonometry...
Click to expand...


Really? ummm..

(-100)(100)+(200)(300)= 50,000



cos= 50000/√(-100^2+100^2)+√(200^2+300^2)

and then you get the inverse of cosine with the answer right?
 
Robertinho! said:
Really? ummm..

(-100)(100)+(200)(300)= 50,000



cos= 50000/√(-100^2+100^2)+√(200^2+300^2)

and then you get the inverse of cosine with the answer right?
Click to expand...

The dot product is this:

\(\displaystyle \displaystyle \binom{a}{b} \binom{x}{y} = \sqrt{a^2 + b^2}\sqrt{x^2+y^2}\cos\theta\)

Notice that there is no plus anywhere. =)
 
Unknown008 said:
The dot product is this:

\(\displaystyle \displaystyle \binom{a}{b} \binom{x}{y} = \sqrt{a^2 + b^2}\sqrt{x^2+y^2}\cos\theta\)

Notice that there is no plus anywhere. =)
Click to expand...


So you don't use the inverse of cos and you do not divide 50000 / the √a^2+b^2*√x^2+y^2?

and one more question. You don't use the pounds AT ALL? right?
 
Robertinho! said:
So you don't use the inverse of cos and you do not divide 50000 / the √a^2+b^2*√x^2+y^2?

and one more question. You don't use the pounds AT ALL? right?
Click to expand...

No you misunderstood me.

Your working showed this:

cos A = 50000/√(-100^2+100^2)+√(200^2+300^2)

I put A because you cannot have cos alone and let A be the angle you're looking for.

The plus sign between the two square roots should NOT be there. Rearranging the formula I posted gives you this:

cos A = 50000/√(-100^2+100^2)√(200^2+300^2)

Notice that there is no plus sign between the square roots.

Secondly, you didn't take the correct numbers in the square roots. They should be:

cos A = 50000/√(-100^2+200^2)+√(100^2+300^2)

Then you should get the answer.

If you look at the formula I posted, -100 becomes a, 200 becomes b, 100 becomes x and 300 becomes y. Notice that x and y are under the same square root, just as a and b.

I am not using pounds at all, since all the units are the same (pounds) and angle is not in pounds but in degrees.
 
A weight of 500 pounds is supported by two cables that exert forces of -100i + 200j and 100i + 300j respectively. Find the angle between the cables.


\(\displaystyle \cos(\theta) \ = \ \dfrac{a\cdot b}{|a|*|b|}\)

\(\displaystyle \cos(\theta) \ = \ \dfrac{(-100i + 200j)\cdot (100i + 300j)}{\sqrt{(-100)^2+300^2} *\sqrt{(100)^2+200^2} }\)

\(\displaystyle \cos(\theta) \ = \ \dfrac{(-100)(100) + (200)(300)}{50000\sqrt{2}} }\)

\(\displaystyle \cos(\theta) \ = \ \dfrac{1}{\sqrt{2}} }\)

\(\displaystyle \theta \ = \ \dfrac{\pi}{4} }\)
 
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