Vectors- Car on a ramp

1. Dec 30, 2011

heartOFphysic

(see image for question)

I just don't know what this question means!! What is the weight "parallel" to the ramp? Please any help on this would be appreciated. I can resolve vectors, but in this particular question, I just don't quite get what to resolve- it's confusing me a little bit, can someone please clarify this question and assist me?

Thanks!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 30, 2011

JHamm

The weight is $mg$ so you'll need to resolve this vector (which obviously points straight down) into a component which is at 30 degrees to the horizontal (a little hint is that it always involves multiplying by sine or cosine of the angle of the ramp, you can draw out triangles and such but I find it more useful to figure out what you want to happen at 0 and 90 degrees to pick which trig function to use).

3. Dec 30, 2011

heartOFphysic

(see image)
Yes so I get that! and the Hypotenuse is parallel to the ramp. so using Sine= Opp/Hyp

we get sin30= 7200N/Hyp

Hyp= 7200/sin30

so Hypotenuse = 14400N????? so I get double the weight not half the weight!

4. Dec 30, 2011

JHamm

Your triangle is set up wrong, you need one where the right angle is the formed between the ramp and its normal vector and the hypotenuse is the force of gravity. More simply just figure out whether sine or cosine is the appropriate choice and multiply the gravitational force by the trig function you chose of 30 degrees.

5. Dec 30, 2011

heartOFphysic

I'm not quite sure what you mean, I can't picture this triangle at all- apologize for my ignorance!

I know the only way I can arrive at half the weight is by the calculation
sin30 X 7200 = 3200N

or Cos60 X 7200 = 3200N

6. Dec 30, 2011

heartOFphysic

anyone?

7. Dec 30, 2011

cupid.callin

check img

and you know component of any vector (of magnitude a) along any other line making angle b with it is a cos(b)

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