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05_InstSolManual_PC

# 05_InstSolManual_PC - USING NEWTON'S LAWS 5 v v v EXERCISES...

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USING NEWTON’S LAWS E XERCISES Section 5.1 Using Newton’s Second Law 12. Newton’s second law for this mass says net 1 2 , F F F ma + = v v v v = where we assume no other significant forces are acting. Since the acceleration and the first force are given, one can solve for the second, 2 2 1 (1.5 kg)(7.3 m/s ) F ma F = v v v = - ˆ ˆ ˆ ˆ ˆ ( cos30 sin30 ) (6.8 N) (2.68 5.48 )N. i j i i j ° ° - + = + This has magnitude 6.10 N and direction 63.9 ° CCW from the x axis. 13. I NTERPRET In this problem, two forces are exerted on the object of interest and produce an acceleration. With the mass of the object and one force given, we are asked to find the other force. D EVELOP Newton’s second law for this mass says net 1 2 , F F F ma = + = r r r r where we assume no other significant forces are acting. The second force is given by 2 1 F ma F = - r r r E VALUATE Substituting the expressions given in the problem statement, we obtain 2 2 2 1 ˆ ˆ ˆ ˆ ˆ ˆ (3.1 kg)[(0.91 m/s ) (0.27 m/s ) ] [( 1.2 N) (2.5 N) ] (4.02 N) (1.66 N) F ma F i j i j i j = - = - - - - = + r r r A SSESS To show that the answer is correct, let’s add the two forces together. The net force is net 1 2 ˆ ˆ ˆ ˆ [( 1.2 N) (2.5 N) ] [(4.02 N) (1.66 N) ] ˆ ˆ ˆ ˆ (2.82 N) ( 0.84 N) (3.1)[(0.91 N) ( 0.27 N) ] F F F i j i j i j i j = + = - - + + = + - = + - r r r From the expression, it is clear that net F r points in the same direction as . a r 14. The horizontal forces on the barge are the two tensions and the resistance of the water, as shown in the figure. The net force is in the x direction, so res 2 cos25 , x T F ma ° - = v v since 1 2 . T T T = = (a) If res 0, 2(1100 N )cos25 1.99 kN. x a F = = ° = v (b) If 2 2 res 0.16 m/s , 1.99 kN (3700 kg)(0.16 m/s ) 1.40 kN. x a F = = - = v 15. I NTERPRET In this problem, we are asked to find the tilt angle of the table such that the acceleration of an object sliding on the surface of the table is the same as the gravitational acceleration near the surface of the Moon. D EVELOP In Example 5.1 it was shown that the acceleration down an incline is sin . a g θ = This equation is what we shall use to solve for the tilt angle . θ 5.1 5

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5.2 Chapter 5 E VALUATE Given that the gravitational acceleration near the surface of the Moon is 2 1.6 m/s , g = the angle of tilt should be 2 1 1 2 1.6 m/s sin sin 9.40 9.8 m/s a g θ - - = = = ° A SSESS A tilt angle of 9.40 ° would give an acceleration that is the same as that of the Moon’s surface. On the other hand, if we want to simulate the motion near Earth’s surface, then , a g = and 90 . θ = ° The greater the value of a , the larger the tilt angle. 16. The acceleration down a frictionless incline is sin a g θ = (see Example 5.1), and the distance traveled down the incline, starting from rest 0 ( 0) v = at the top 0 ( 0), x = is 2 1 2 . x at = Therefore 2 2 / sin 2(1.3 km)/(9.8 m/s )sin 24 25.5 s. t x g θ = = ° = 17. I NTERPRET In this problem, the physical quantity of interest is the tension force in the cable. To compute the tension, we analyze the motion of the car.
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05_InstSolManual_PC - USING NEWTON'S LAWS 5 v v v EXERCISES...

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