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course Phy 201
10/29 2 pmMore will be added to Class Notes, but here are the questions:
`q001. If the acceleration of an Atwood system with total mass 80 grams is 50 cm/s^2, then:
* How much mass is on each side? Note that this can be reasoned out easily without an complicated analysis, using the same type of reasoning that led us to conclude that a system with 31 g on one side and 30 g on the other accelerates at 1/61 the acceleration of gravity.
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38 g on one side and 42 on the other. The system accelerates at 1/20 of that of gravity roughly.
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* By analyzing the forces on the mass on the 'lighter' side, what is the tension in the string?
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Tension=38 g(50 cm/s^2)= 1900 g*cm/s^2 or 0.0019 N
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* By analyzing the forces on the mass on the 'heavier' side, what is the tension in the string?
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Tension=42 g(50 cm/s^2)=2100 g*cm/s^2 or 0.0021 N
That would be the net force, not the tension. Gravity is also present.
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`q002. A large sweet potato has mass 1188.6 grams. Sweet potatoes sink in water. Suppose that when this sweet potato is suspended as the mass on one side of an Atwood machine, but immersed in water, a mass of 100 grams on the other side is required to balance it.
* Sketch the forces on that system and describe your sketch.
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Drawing the diagram, I have a sweet potato submerged in water and the 100g mass on the other side. From each, I have an mg vector to denote the force of gravity. I also have tension forces on each side of the string.
The weight vector for the greater mass should be about 12 times as long as for the lesser mass.
Tension forces should be equal, upward on each object, and such that the acceleration has the same magnitude on both sides.
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* Would the sweet potato, if reshaped without changing its volume, fit into that 1-liter container?
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Yes because as we described in class there is a +/- 10% margin that would account for the 188 g extra to fit.
The volume of the container and the weight indicated by the balance are highly accurate, both to within at most +- 1/10 of 1%. There could be enough friction in the pulley to make a difference corresponding to 1 gram.
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* What if the required balancing mass was 200 grams?
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No, it would not fit because it would make the % margin too great.
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`q003. The balloon rose to the ceiling in about 1.5 seconds, 2 seconds, 2.5 seconds and 5 seconds when 1, 2, 3 and 4 paperclips, respectively, were attached. It fell to the floor in about 6 seconds when 5 paperclips were attached. Assume the displacement to have been the same in each case. The buoyant force results from the fact that air pressure decreases as altitude changes, which results in more force from the air pressure on the bottom of the balloon than on the top. The pressure in the room changes at a very nearly constant rate with respect to altitude, so the buoyant force can be assumed to remain constant throughout the room.
* Assuming uniform acceleration in each case, is a graph of acceleration vs. number of clips linear?
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No, it is not linear, the graph looks like a defunct curve, sort of.
You refer to a graph, but you need to state what is graphed vs. what.
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* Do your results indicate the presence of a force other than the gravitational and buoyant forces acting on the balloon?
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Yes? Since the graph is not linear, then it indicates the presence of another force.
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`q004. When a certain object coasts up an incline its acceleration has magnitude 100 cm/s^2 and is directed down the incline. When it coasts down the incline its acceleration is 50 cm/s^2 and is directed up the incline. Only gravitational, normal and frictional forces are present.
* Sketch a figure depicting the forces on the object as it coasts up, and as it coasts down the incline. Describe your sketch.
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Sketching the object the coasting up the incline, I first put in my mg vector and my equal/opposite Fnorm vector; however it is not truly equal and opposite as it must be perpendicular to the x axis. Coasting up the incline, the friction vector is a small vector pointing downward. Coasting up the incline, the friction vector is small like the previous vector. The mag of acceleration vector down the ramp is twice as large as the acceleration up the ramp.
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* Are the magnitudes of the force vector depicted in your sketch consistent with the given accelerations? If not, make another sketch and adjust the vectors as necessary. Then describe why you think your sketch is a reasonable representation of the system.
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I think my vectors are reasonably accurate. The one vector is half the size of the 100 cm/s^2 vector.
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* From the given information you can determine the coefficient of friction. You may assume that the normal force varies little in magnitude from the weight of the object. What is the coefficient of friction?
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.05 (acceleration/acceleration of gravity). The change in acceleration is 50 cm/s^2 and gravity is 980 cm/s^2
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* Having determined the coefficient of friction, you can also determine the slope of the incline. For simplicity you can assume that since the slope is small, the magnitude of the weight component parallel to the incline is equal to the slope multiplied by the weight of the object. What do you get?
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We are not given weight or an angle, so I’m unsure as to how to pursue this question,
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`q005. If parts of the preceding problem gave you trouble, consider an object on an incline with slope .05 and coefficient of friction .03. You can again assume that since the slope is small, the magnitude of the weight component parallel to the incline is equal to the slope multiplied by the weight of the object, and also that the normal force does not differ significantly in magnitude from the object's weight. Let m stand for the mass of the object, g for the acceleration of gravity.
In terms of m and g:
* What is the weight of the object?
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m*g
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* What is the magnitude of its weight parallel to the incline?
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mg cos (theta)
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* What is the magnitude of the normal force?
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equal and opposite mg, only it must be perpendicular to the x axis.
the normal force in this case is equal and opposite to the component of the weight which is perpendicular to the incline
that component is m g sin(theta)
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* What is the magnitude of the frictional force?
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mu * mg
mu * normal force; normal force isn't mg in this case
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* What is the magnitude of the net force when the object coasts up the incline?
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mg+Ffriction=Fnet
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* What is the magnitude of the net force when the object coasts down the incline?
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Fnet=mg-Ffriction
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* What therefore are the object's accelerations up, and down, the incline?
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a=mg-Ffriction and a= mg+Ffriction
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* What are those accelerations in cm/s^2?
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-100 cm/s^2 and 50 cm/s^2
this would not be the case for the present object, which is on slope .05 with coefficient of friction .03
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