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course Phy 241
November 7 around 9:40am. I submitted this lab a while back and it never came up in my portfolio, so I'm just re-submitting it.
`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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(x/80g)(g)=a
(x/80g)=(50cm/s^2/980cm/s^2)
x=4.1g
(m2-m1=4.1g) and (m2+m1=80g) solved simultaneously
(m2=80-m1), (80-m1-m1=4.1g), (m1=37.95g)
(m2=80g-37.95g), (m2=42.05g)
There’s probably an easier way, but I thought this was the easiest to reason out, personally.
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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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Lighter side is 37.95g
Fnet+(m1*g)=T
Fnet=(80g)*(50cm/s^2)=4000g*cm/s^2
4000+(37.95*980)=T
41191N=T
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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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Heavier side is 42.05g
Fnet-(m_2*g)=-T
-Fnet+(m_2*g)=T
-4000+(42.05*980)=T
37209N=T
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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 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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Forces: upward buoyant force on potato, downward gravitational force on potato and object on other side
The gravitational force of the sweet potato is about 12 times the length of the gravitational force of the object on the other side. (Potato’s mass is about 1200g while the object is 100g). The upward buoyant force on the potato has a length that should be equal to the difference of the 2 gravitational forces.
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* Would the sweet potato, if reshaped without changing its volume, fit into that 1-liter container?
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No, because it has too much mass and the density makes a difference.
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* What if the required balancing mass was 200 grams?
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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, because between 1-3 paperclips, the time intervals increase by 0.5seconds until you reach the 4th paperclip, then the time interval from jumps from 2.5 seconds to 5 seconds, which is doubled. This, therefore, would make the graph linear UNTIL the 4th paperclip is added, then it would change.
I agree that acceleration isn't linear. However the acceleration is neither proportional to the time interval, nor inversely proportional to it. For a given distance the acceleration from rest is inversely proportional to the square of the time interval.
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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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No, there IS other forces (downward force from gas in balloon, downward force from some air resistance, and maybe some atmospheric pressure) but looking at the results above, there’s only gravitational and buoyant forces.
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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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Coasts UP: frictional force is downward (opposite to motion), Fnorm is perp to incline, and gravitational force is straight down from the object itself.
Coasts DOWN: frictional force is upward (opposite of motion), Fnorm is perp to incline, and gravitational force is straight down from the object itself.
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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, because I drew my sketch accurately and therefore the magnitudes are consistent with the accelerations, they’re proportional.
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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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Ffric=(100cm/s^2-50cm/s^2)=25
25/(980cm/s^2)=mu
0.025=mu
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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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(wt_parallel)/(wt)=slope
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Friction acts up in one case, down in the other so the difference in forces is twice the magnitude of the frictional force.
The difference in forces is (100 cm/s^2 - 50 cm/s^2) * mass = 50 cm/s^2 * mass, so frictional force is 25 cm/s^2 * mass.
This is mu * mass * g, where mu = .025.
The parallel gravitational component is 25 cm/s^2 * mass less one force and 25 cm/s^2 * mass greater than the other. It follows that the parallel gravitational component is 75 cm/s^2 * mass, and the slope is .075.
`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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[wt_parallel=slope*(wt)]
[Fnorm=wt, mg=wt, Fnorm=mg]
wt=mg
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* What is the magnitude of its weight parallel to the incline?
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wt_parallel=slope(mg)
wt_parallel=0.05(mg)
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* What is the magnitude of the normal force?
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Fnorm=mg
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* What is the magnitude of the frictional force?
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Ffric=mu*mg
Ffric=0.03*mg
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* What is the magnitude of the net force when the object coasts up the incline?
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UP: -0.03mg [Fnet=mu*mg]
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* What is the magnitude of the net force when the object coasts down the incline?
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DOWN: 0.03mg
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* What therefore are the object's accelerations up, and down, the incline?
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[a=Fnet/m]
UP: -0.03mg/m=(-0.03g)
DOWN: 0.03mg/m=(0.03g)
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* What are those accelerations in cm/s^2?
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UP: -29.4cm/s^2 [-0.03(980cm/s^2)]
DOWN: 29.4cm/s^2 [0.03(980cm/s^2)]
Check class notes on this last one. The accelerations are .02 g and .08 g.
&#Good responses. See my notes and let me know if you have questions. &#
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