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course PHY 231
10/28 3:40 pm
Here are some questions related to the 101025 class. Notes will be completed soon.
`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 a 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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36, 44
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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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T-m1g=80*50----> T= 40000 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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m2g-T=80850-----> T= 40000 N and T is the same on both sides as it should be.
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We can reason that since 50 cm/s^2 is about 1/20 the acceleration of gravity, the difference in the masses should be 1/20 the total mass. That would be a 4 gram difference, implying masses of 38 grams and 42 grams.
`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 one 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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The mass has two forces acting on it: gravity down and tension up. The sweet potato has several: tension up, gravity down, buoyant force up, and water pressure down.
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water pressure acts in all directions, and the net effect of water pressure is the buoyant force itself.
Would the sweet potato, if reshaped without changing its volume, fit into that 1-liter container?
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Well, here's how I see it: 1188.6-100= 1088.6 g. So the sweet potato is slightly more dense than water, which means that if d_sp=d_h20 then g_sw/1L=g_420/1L and this is not true. The volume of the sweet potato would have to be about 1.08 L to balance the water.
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What if the required balancing mass was 200 grams?
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Then the potato would be less dense and would fit into the container.
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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, if it were linear then each clip would add the same amount of travel time and it doesn't because the times are not linear, +3clips = +3.5s.
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travel time is not proportional to acceleration, nor is it inversely proportional
How would travel time be expected to vary with net force if acceleration vs. number of clips was linear?
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 because g is effectively constant over the small distance travelled so something else is adding nonlinearly to the downward force with each clip.
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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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Both ways have normal and weight forces acting in the same directions both times, that is perp for norm and perp/parallel for gravity with parallel down incline. Going up there is friction also acting down the ramp and going down friction is acting up the slope.
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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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Yes, gravity is the largest force here and the normal and then frictional forces are fractions of that. Since friction acts with gravity when going up the incline the net force is twice as long as when friction acts against gravity.
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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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Well we have two eq and two unknowns. The system is: sl+µ=0.1 and sl-µ=-0.05, so µ =0.075
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Good. The complete solution follows immediatly from yours: mu = .075 and slope = .025
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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From the above we see the slope must be 0.025
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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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mg
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What is the magnitude of its weight parallel to the incline?
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slope*mg = 0.05mg
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What is the magnitude of the normal force?
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mg
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What is the magnitude of the frictional force?
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µmg =0.03mg
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What is the magnitude of the net force when the object coasts up the incline?
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0.08mg
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What is the magnitude of the net force when the object coasts down the incline?
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0.02mg
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What therefore are the object's accelerations up, and down, the incline?
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0.08g and 0.02g
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What are those accelerations in cm/s^2?
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About 80 cm/s^2 and 20 cm/s^2
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Good, but see my notes, especially on the first question.
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