Question 117104
This was previously answered as the posting to problem 117069 (answer 85147). To save you the
trouble of looking that answer up, here it is again:
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In this forum it's pretty hard to do drawings that support geometric problems such as this one.
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So, let's just say that for a weight on a ramp that goes up at an angle {{{alpha}}} the 
portion of the weight W that acts perpendicular to the ramp is given by the equation:
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{{{W[n] = W*cos(alpha)}}}
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and the portion of the weight W that acts parallel to and down the ramp is given by the 
equation:
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{{{W[t] = W*sin(alpha)}}}
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In this problem, the weight W is the weight of the SUV which is given as 6000 lbs and the
angle {{{alpha}}} of the ramp is 25 degrees.
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The winch will have to generate enough force to overcome the portion of the weight of the
SUV that acts parallel to and down the ramp. [This assumes the rolling friction of the vehicle
is negligible.] So the winch will have to produce an amount of force equal to:
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{{{W[t] = W*sin(alpha) = 6000*sin(25) = 6000* 0.422618261 = 2535.71}}}
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Once the vehicle is in motion the winch will have to pull with a force of 2535.71 lbs 
parallel to the ramp to keep the SUV rolling up the ramp at a steady rate of movement.
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Hope this helps you to understand the problem a little better. If you expect to do many of
these problems, you might want to memorize these two "ramp" equations.
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