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Am I on the right track when I say: x=2000cos 30? Then what?

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- Thread starter Rainydays253
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- #1

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Am I on the right track when I say: x=2000cos 30? Then what?

- #2

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Well the basic idea is good but I don't know how you came up with cos. If you look at a scetch you should be abel to find the right function for the force perralel to the slope. And just a small hint when your searching for a force it's usually a good idea to use weight instead of mass (I'm not very good at non-metric units soo I don't really know wheater lb steands for mass or weight).

Hope this helps.

PS. Maybe this thread would be more sutible for the Science Education Zone.

Hope this helps.

PS. Maybe this thread would be more sutible for the Science Education Zone.

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- #3

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Hmm...what is the coefficient of friction between the tires and the incline?

Assuming the ramp is frictionless, your "force" must contain a component parallel to the ramp, with magnitude:

[tex] F = \left( {2000 \, {\text{lb}}} \right)\sin \frac{\pi }{6} = 1000\,{\text{lb}} [/tex]

Basically, gravity will exert**1000 lb** down the ramp. To keep the car from rolling down, you must exert a force of **1000 lb** UP the ramp (parallel to the ramp, antiparallel to the component of weight parallel to the ramp).

However, if you*do* have friction, then to prevent __rolling__ down the ramp (NOT skidding!), you must find the coefficient of 'static' friction between the tires and the ramp. Now, let [itex] \mu _S [/itex] represent this value. Thus, the force needed to prevent the car from rolling down is:

[tex] F = 2000\,{\text{lb}}\left( {\sin \frac{\pi }{6} - \mu _S \cos \frac{\pi }{6}} \right) = 1000\,{\text{lb}}\left( {1 - \mu _S \sqrt 3 } \right) [/tex]

**In the direction antiparallel to the weight component parallel to the ramp.

Anyway, those are always helpful (Unless you're in a pinch for time!...or the problem is just basic).

Also "lb" stands for pounds, which are a customary unit of*weight*, I believe, according to

http://dictionary.reference.com/search?q=lb

http://dictionary.reference.com/search?q=pound

Assuming the ramp is frictionless, your "force" must contain a component parallel to the ramp, with magnitude:

[tex] F = \left( {2000 \, {\text{lb}}} \right)\sin \frac{\pi }{6} = 1000\,{\text{lb}} [/tex]

Basically, gravity will exert

However, if you

[tex] F = 2000\,{\text{lb}}\left( {\sin \frac{\pi }{6} - \mu _S \cos \frac{\pi }{6}} \right) = 1000\,{\text{lb}}\left( {1 - \mu _S \sqrt 3 } \right) [/tex]

**In the direction antiparallel to the weight component parallel to the ramp.

What basic idea? Also, by "sketch", do you mean a free-body diagram?Well the basic idea is good but I don't know how you came up with cos. If you look at a scetch you should be abel to find the right function for the force perralel to the slope. And just a small hint when your searching for a force it's usually a good idea to use weight instead of mass (I'm not very good at non-metric units soo I don't really know wheater lb steands for mass or weight).

Anyway, those are always helpful (Unless you're in a pinch for time!...or the problem is just basic).

Also "lb" stands for pounds, which are a customary unit of

http://dictionary.reference.com/search?q=lb

http://dictionary.reference.com/search?q=pound

Last edited:

- #4

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Thanks for the explination.

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