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course PHY 241
10/17/20113:40 am
I must apologize for how unbelievably late this is, I am finally mostly caught up with my work, so this shouldn't happen with any new assignments." "`q001. For each of the given objects on the various inclines estimate, based on a sketch as opposed to a formula, the parallel and perpendicular components of the object's weight as a percent of its weight. Use your estimated percents to find the component of each weight parallel to the incline, and perpendicular to the incline.
A car weighing 20 000 Newtons on an incline making angle 12 degrees with horizontal.
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parallel to incline: 19563 newtons
perpendicular to incline: 4158 newtons
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@& On a flat incline the angle would be 0 degrees and all the force would be perpendicular to the incline.
At 12 degrees the perpendicular component of the force would still be close to 20 000 N.
These results look at though they were calculated rather than estimated. You need to carefully relate this situation to your picture, then estimate, and if you do use the sine and cosine you need to be sure you are using the definitions correctly.*@
A hotwheels car of weight 80 000 dynes on an incline whose angle with horizontal is 20 degrees.
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parallel to incline: 75175.4 dynes
perpendicular to incline: 27361.6 dynes
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A block of weight 30 pounds on a 37 degree incline.
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parallel to incline: 23.97 pounds
perpendicular to incline: 18.05 pounds
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@& For inclines at less than 45 degrees the parallel component will be less than the perpendicular.
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`q002. If the car in the first question experiences a frictional force which is 2% of the perpendicular component of its weight, then what is the magnitude of the frictional force?
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0.02 * 4158 newtons = 83.16 newtons
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If the car is coasting downhill what is the sum of the parallel component of its weight and the frictional force?
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19563 + (-83.16) = 19479.8 newtons
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If the car is coasting uphill what is the sum of the parallel component of its weight and the frictional force?
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???
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@& The parallel component of the gravitational force is much smaller than th perpendicular at this angle.*@
`q003. If the hotwheels car in the first question is attached by a light thread to a washer weighing 20 000 dynes and suspended over a pulley at the lower end of the ramp, then if friction is ignored what is the net force acting in the direction down the ramp?
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95175.4 dynes
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Answer the same question assuming that the washer is suspended from a pulley at the top of the ramp.
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net force down the ramp: 55175.4 dynes
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`q004. How much frictional force would it take to hold he block in the first question stationary on the incline?
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23.97 pounds
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What is this force as a percent of the weight of the block?
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79.9%
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"
@& You've mixed up the parallel and perpendicular components of the weight on some problems, but not on all.
I've inserted notes on some but not all of the questions on which you appear to have done so.
&#Please see my notes and, unless my notes indicate that revision is optional, submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.
If my notes indicate that revision is optional, use your own judgement as to whether a revision will benefit you. &#
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