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#1 2009-06-02 05:06:43
- EastEnder905
- Member
- Registered: 2009-06-02
- Posts: 2
Tension/Equilibrium
Hello all, I just registered with this site because I could use a little help with some math questions. The question that I ask of you to help me with is going to serve me as a guideline for other questions very similar to this that are all over my text book, so all help is welcome and appreciated.
Q:
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#2 2009-06-02 05:41:45
- mathsyperson
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- Registered: 2005-06-22
- Posts: 4,900
Re: Tension/Equilibrium
Make equations in terms of the vertical and horizontal components of each force.
The weight's force is entirely vertical and downwards.
AC has no vertical component.
The 150lb weight has a vertical component of 150sin30 = 75lb, upwards.
AB has a vertical component of Bsin60, where B is the tension of AB.
The overall equation for vertical components is therefore
Bsin60 + 75 = 200
Similarly, the equation for horizontal components is
Bcos60 + C = 150cos30
You can now find B from the first equation, then use that value with the second to find C.
(This is generally how you would solve something like this, although you don't have to use vertical and horizontal components. For example, if a slope is involved in a question, it might be better to use components that are parallel and perpendicular to it instead.)
Why did the vector cross the road?
It wanted to be normal.
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#3 2009-06-02 08:54:22
- EastEnder905
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- Registered: 2009-06-02
- Posts: 2
Re: Tension/Equilibrium
I am slightly confused with the answer you gave me. How would you rearrange to first formula to get B?
B=(200-75)/sin60?
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#4 2009-06-02 08:55:42
- mathsyperson
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- Registered: 2005-06-22
- Posts: 4,900
Re: Tension/Equilibrium
Yep!
Why did the vector cross the road?
It wanted to be normal.
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