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course Phy 201
Oct 10, 2011 at 4:20We're just supposed to read and understand the Introductory Problem Set 5 right? Submitting it is not required?
@& Right.*@
As discussed in class you should master Introductory Problem Set 5 before next Monday's class.
You should also submit your answers to the following questions:
`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 Component
25% of 20 000 Newtons = 5 000 Newtons
Perpendicular Component
90% 0f 20 000 Newtons = 18 000 Newtons
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A hotwheels car of weight 80 000 dynes on an incline whose angle with horizontal is 20 degrees.
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Parallel Component
30% of 80 000 dynes = 24 000 dynes
Perpendicular Component
85% of 80 000 dynes = 68 000 dynes
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A block of weight 30 pounds on a 37 degree incline.
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Parallel Component
65% of 30 Ibs = 19.5 Ibs
Perpendicular Component
80% of 30 Ibs = 24 Ibs
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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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frictional force = 2% of the perpendiclar component
= 2% of 18 000 Newtons
= 360 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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parallel component + (-)frictional force
5 000 Newtons - 360 Newtons = 4640 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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parallel compontent + frictional force
5 000 Newtons + 360 Newtons = 5360 Newtons
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`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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F_net = m a
= (20 000 dynes + 80 000 dynes) (9.8 m/s^2)
= 100 000 dynes * 9.8 m/s^2
= 980 000 dynes m / s^2
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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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F_net = m a
= (-20 000 dynes + 80 000 dynes) (9.8 m/s^2)
= 60 000 dynes * 9.8 m/s^2
= 588 000 dynes m / s^2
@& Your results would be correct if the ramp was vertical.
The question didn't really specify which of the ramps, but the answer would differ from one ramp to the other.
Instead of the 80 000 N you would use the component of the weight parallel to the particular ramp.
I don't think this will give you any trouble.*@
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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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perpendicular component is 24 Ibs of the original 30 Ibs
so it would take 6 Ibs of frictional force?
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What is this force as a percent of the weight of the block?
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30 - 24 = 6
6 / 30 = 0.2 ---> 20 % frictional force
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@& Again there are three inclines, and the way the question is posed isn't clear. Especially since no block was mentioned in that problem.
However if the frictional force is 25% of the perpendicular component, then your answers make very good sense.*@
&#Good responses. See my notes and let me know if you have questions. &#