HW3 - Stage 0 Solutions: 5.1. Model: We can assume that the...

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Stage 0 Solutions: 5.1. Model: We can assume that the ring is a single massless particle in static equilibrium. Visualize: Solve: Written in component form, Newton’s first law is () net 1 2 3 0 N xxxx x FF T T T =Σ = + + = ( ) net 1 2 3 yyyy y T T T + + = Evaluating the components of the force vectors from the free-body diagram: 1 x T =− 1 T T 2 x = 0 N 33 cos30 x TT = ° T 1 y = 0 N 2 y 2 = sin30 y = −° Using Newton’s first law: 13 cos30 0 N −+ ° = 23 sin30 0 N = Rearranging: ( ) cos30 100 N 0.8666 86.7 N = = ( )( ) sin30 100 N 0.5 50.0 N = = Assess: Since acts closer to the x -axis than to the y -axis, it makes sense that . 3 T r 12 > 5.5. Visualize: Please refer to the Figure Ex5.5. Solve: Applying Newton’s second law to the diagram on the left, net 2 4 N 2 N 1.0 m/s 2 kg x x F a m == = ( ) net 2 3 N 3 N 0 m/s 2 kg y y F a m = For the diagram on the right: net 2 4 N 2 N 1.0 m/s 2 kg x x F a m = ( ) net 2 3 N 1 N 2 N 0 m/s 2 kg y y F a m −− =
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5.7. Visualize: Please refer to Figure Ex5.7. Solve: We can use the constant slopes of the three segments of the graph to calculate the three accelerations. For t between 0 s and 3 s, 2 12 m/s 0 s 4 m/s 3 s x x v a t Δ− == = Δ For t between 3 s and 6 s, Δ v x = 0 m/s, so a x = 0 m/s 2 . For t between 6 s and 8 s, 2 0 m/s 12 m/s 6 m/s 2 s x x v a t = Δ From Newton’s second law, at t = 1 s we have F net = ma x = (2.0 kg)(4 m/s 2 ) = 8 N At t = 4 s, a x = 0 m/s 2 , so F net = 0 N. At t = 7 s, F net = ma x = (2.0 kg)(–6.0 m/s 2 ) = –12 N Assess: The magnitudes of the forces look reasonable, given the small mass of the object. The positive and negative signs are appropriate for an object first speeding up, then slowing down. 5.9. Model: We assume that the box is a particle being pulled in a straight line. Since the ice is frictionless, the tension in the rope is the only horizontal force. Visualize: Solve: (a) Since the box is at rest, a x = 0 m/s 2 , and the net force on the box must be zero. Therefore, according to Newton’s first law, the tension in the rope must be zero. (b) For this situation again, a x = 0 m/s 2 , so . net 0 N FT (c) Here, the velocity of the box is irrelevant, since only a change in velocity requires a nonzero net force. Since a x = 5.0 m/s 2 , ( )( ) 2 net 50 kg 5.0 m/s 250 N x m a = = Assess: For parts (a) and (b), the zero acceleration immediately implies that the rope is exerting no horizontal force on the box. For part (c), the 250 N force (the equivalent of about half the weight of a small person) seems reasonable to accelerate a box of this mass at 5.0 m/s 2 . 5.13. Model: We assume that the passenger is a particle subject to two vertical forces: the downward pull of gravity and the upward push of the elevator floor. We can use one-dimensional kinematics and Equation 5.10.
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Visualize:
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Solve: (a) The apparent weight is () 2 app 0 1 1 60 kg 9.8 m/s 590 N y a ww w m g gg ⎛⎞ =+=+ = = = ⎜⎟ ⎝⎠ (b) The elevator speeds up from v 0y = 0 m/s to its cruising speed at v y = 10 m/s. We need its acceleration before we can find the apparent weight: 2 10 m/s 0 m/s 2.5 m/s 4.0 s y v a t Δ− == = Δ The passenger’s apparent weight is 2 app 2 2.5 m/s 1 590 N 1 9.8 m/s y a g =+= + = (590 N)(1.26) = 740 N (c) The passenger is no longer accelerating since the elevator has reached its cruising speed. Thus,
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This note was uploaded on 06/20/2008 for the course PHYS 125 taught by Professor Mutchler during the Spring '08 term at Rice.

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HW3 - Stage 0 Solutions: 5.1. Model: We can assume that the...

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