course Phy 231
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18:16:57 Query set 3 #'s 13-14 If an object of mass m1 rests on a frictionless tabletop and a mass m2 hangs over a good pulley by a string attached to the first object, then what forces act on the two-mass system and what is the net force on the system? What would be the acceleration of the system? How much would gravitational PE change if the hanging mass descended a distance `dy?
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RESPONSE --> Fnet on the system is simply m2 * g a = Fnet/(m1+m2) PEgrav = -F'dy (loss of PE since m2 is lowering); opposite and equal the kE of the system
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18:18:19 ** The net force on the system is the force of gravity on the suspended weight: Fnet = m2*9.8m/s/s Gravity also acts on m1 which is balanced by force of table on m1, so this force makes no contribution to Fnet. Acceleration=net force/total mass = 9.8 m/s^2 * m2 / (m1+m2). If the mass m2 descends distance `dy then gravitational PE decreases by - m2 g * `dy. COMMON MISCONCEPTIONS AND INSTRUCTOR COMMENTS: The forces acting on the system are the forces which keep the mass on the table, the tension in the string joining the two masses, and the weight of the suspended mass. The net force should be the suspended mass * accel due to gravity + Tension. INSTRUCTOR COMMENT: String tension shouldn't be counted among the forces contributing to the net force on the system. The string tension is internal to the two-mass system. It doesn't act on the system but within the system. Net force is therefore suspended mass * accel due to gravity only 'The forces which keep the mass on the table' is too vague and probably not appropriate in any case. Gravity pulls down, slightly bending the table, which response with an elastic force that exactly balances the gravitational force. **
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RESPONSE --> I could have replaced F in parts 2 and 3 with m2*g
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18:21:35 How would friction change your answers to the preceding question?
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RESPONSE --> Friction would cause the net force to be reduced from its original amount of m2*g by an amount equal to the product of the coefficient of friction and the normal force (equal to and opposite to the weight of m1). A lower net force would create a smaller acceleration (a = Fnet/(m1+m2)) The PE change would remain the same, however; the kE change will not equal the PE change, the difference being the work done by the system against friction.
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18:21:49 **Friction would act to oppose the motion of the mass m1 as it slides across the table, so the net force would be m2 * g - frictional resistance. **
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18:23:31 Explain how you use a graph of force vs. stretch for a rubber band to determine the elastic potential energy stored at a given stretch.
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RESPONSE --> Since work = F*'ds, we can approximate the elastic PE (work that could be done) by taking the area under the curve for the given stretch interval; the most exact way would involve taking the definite integral of the function for force vs. stretch between the desired stretch amounts.
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18:23:47 ** If we ignore thermal effects, which you should note are in fact significant with rubber bands and cannot in practice be ignored if we want very accurate results, PE is the work required to stretch the rubber band. This work is the sum of all F * `ds contributions from small increments `ds from the initial to the final position. These contributions are represented by the areas of narrow trapezoids on a graph of F vs. stretch. As the trapezoids get thinner and thinner, the total area of these trapezoids approaches, the area under the curve between the two stretches. So the PE stored is the area under the graph of force vs. stretch. **
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18:25:39 STUDENT QUESTIONS: Does the slope of the F vs stretch graph represent something? Does the area under the curve represent the work done? If so, is it work done BY or work done ON the rbber bands?
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RESPONSE --> The slope represents the rate at which the force changes with respect to the amount of stretch. The area under the curve is the work done on the rubber band to cause it to stretch (and thus store PE), but also the work that will be done when the band is released, when we ignore the significant thermal energy loss.
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18:26:29 ** Slope isn't directly related to any physical quantity. The area is indeed with work done (work is integral of force with respect to displacement). If the rubber band pulls against an object as is returns to equilibrium then the force it exerts is in the direction of motion and it therefore does positive work on the object as the object does negative work on it. If an object stretches the rubber band then it exerts a force on the rubber band in the direction of the rubber band's displacement, and the object does positive work on the rubber band, while the rubber band does negative work on it. **
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RESPONSE --> OK
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18:28:14 Query rubber bands supporting mass (experiment 11) When rubber bands support a mass, how can you determine the components of the force of one of the rubber bands?
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RESPONSE --> This experiment was not assigned, but I would assume that you could use the angle between the rubber band and the mass it was supporting (which one would assume to be in a downward direction), then finding the components.
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18:28:21 ** The length of the rubber band determines its force, which can be read from the calibration graph. The angle of the rubber bands can be used with the force to find x and y components of each, using standard vector methods. **
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18:28:47 When rubber bands support a mass, how do the horizontal components of their forces compare?
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RESPONSE --> They should be opposite and equal since the system is in equilibrium and there are no other horizontal forces.
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18:29:04 ** If a mass is supported in a stationary position by two rubber bands, then the mass is in equilibrium. It follows that the net force on the mass is zero (if there was a nonzero net force it would have to be accelerating at rate a = F / m). So the net force in any direction must be zero. In particular the net horizontal force must be zero. In general both rubber bands will be pulling at nonzero angles with respect to vertical, so both will have both vertical and horizontal components. In the usual situation the horizontal components are the only forces exerted in the horizontal direction. Since the net force in this direction is zero, the horizontal components must be equal and opposite. The vertical forces acting on the object include the vertical components of the rubber band forces and the weight of the object. The total of the vertical components of the rubber band force is therefore equal and opposite to the downward pull of gravity. **
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18:29:46 When rubber bands support a masss, what can be said about the vertical components of the forces exerted and the force exerted by gravity on the mass?
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RESPONSE --> the vertical components of the forces exerted must be equal to and opposite the force exerted by gravity on the mass since the system is in equilibrium.
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18:29:50 ** The vertical forces must add up to zero, since the object isn't accelerating in the horizontal direction. The vertical components of the forces exerted by the supporting rubber bands will be equal and opposite to the force exerted by gravity. **
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18:29:53 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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