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ch05 - 3 We apply Newton's second law(Eq 5-1 or...

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3. We apply Newton’s second law (Eq. 5-1 or, equivalently, Eq. 5-2). The net force applied on the chopping block is & & & F F F net = + 1 2 , where the vector addition is done using unit-vector notation. The acceleration of the block is given by & & & a F F m = + 1 2 d i / . (a) In the first case ( ) ( ) ( ) ( ) 1 2 ˆ ˆ ˆ ˆ 3.0N i 4.0N j 3.0N i 4.0N j 0 F F ª º ª º + = + + + = ¬ ¼ ¬ ¼ & & so & a = 0 . (b) In the second case, the acceleration & a equals ( ) ( ) ( ) ( ) ( ) ( ) 2 1 2 ˆ ˆ ˆ ˆ 3.0N i 4.0N j 3.0N i 4.0N j ˆ (4.0m/s )j. 2.0kg F F m + + + + = = & & (c) In this final situation, & a is ( ) ( ) ( ) ( ) ( ) ( ) 2 1 2 ˆ ˆ ˆ ˆ 3.0N i 4.0N j 3.0N i 4.0N j ˆ (3.0m/s )i. 2.0 kg F F m + + + + = = & &

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4. The net force applied on the chopping block is & & & & F F F F net = + + 1 2 3 , where the vector addition is done using unit-vector notation. The acceleration of the block is given by & & & & a F F F m = + + 1 2 3 d i / . (a) The forces (in newtons) exerted by the three astronauts can be expressed in unit- vector notation as follows: ( ) ( ) ( ) ( ) ( ) 1 2 3 ˆ ˆ ˆ ˆ 32 cos 30 i sin 30 27.7 i+16 j j ˆ ˆ ˆ 55 cos 0 i sin 0 55i j ˆ ˆ ˆ ˆ 41 cos 60 i sin 60 20.5i 35.5 j. j F F F = ° + ° = = ° + ° = = ° + ° = & & & The resultant acceleration of the asteroid of mass m = 120 kg is therefore ( ) ( ) ( ) 2 2 ˆ ˆ ˆ ˆ ˆ 27.7i 16 j 55i 20.5i 35.5j ˆ ˆ (0.86m/s )i (0.16m/s )j . 120 a + + + = = & (b) The magnitude of the acceleration vector is & a a a x y = + = + = 2 2 2 2 2 086 016 088 . . . . b g m / s (c) The vector & a makes an angle θ with the + x axis, where θ = F H G I K J = F H G I K J = − ° tan tan . . . 1 1 016 086 11 a a y x
5. We denote the two forces & & F F 1 2 and . According to Newton’s second law, & & & & & & F F ma F ma F 1 2 2 1 + = = so , . (a) In unit vector notation & F 1 20 0 = . \$ N i b g and ( ) ( ) ( ) ( ) 2 2 2 2 ˆ ˆ ˆ ˆ 12.0 sin 30.0 m/s i 12.0 cos 30.0 m/s 6.00 m/s i 10.4m/s j. j a = − ° ° = − & Therefore, ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 ˆ ˆ ˆ 2.00kg 6.00 m/s i 2.00 kg 10.4 m/s j 20.0 N i ˆ ˆ 32.0 N i 20.8 N j. F = + = − & (b) The magnitude of & F 2 is 2 2 2 2 2 2 2 | | ( 32.0) ( 20.8) 38.2 N. x y F F F = + = + − = & (c) The angle that & F 2 makes with the positive x axis is found from tan θ = ( F 2 y / F 2 x ) = [(–20.8)/(–32.0)] = 0.656. Consequently, the angle is either 33.0° or 33.0° + 180° = 213°. Since both the x and y components are negative, the correct result is 213°. An alternative answer is 213 360 147 ° − ° = − ° .

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6. We note that ma = (–16 N) i ^ + (12 N) j ^ . With the other forces as specified in the problem, then Newton’s second law gives the third force as F 3 = ma F 1 F 2 =(–34 N) i ^ (12 N) j ^ .
7. Since & v = constant, we have & a = 0, which implies & & & & F F F ma net = + = = 1 2 0 . Thus, the other force must be 2 1 ˆ ˆ ( 2 N) i (6 N) j. F F = − = − + & &

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8. From the slope of the graph we find a x = 3.0 m/s 2 . Applying Newton’s second law to the x axis (and taking θ to be the angle between F 1 and F 2 ), we have F 1 + F 2 cos θ = ma x ¡ θ = 56 ° .
9. (a) – (c) In all three cases the scale is not accelerating, which means that the two cords exert forces of equal magnitude on it. The scale reads the magnitude of either of these forces. In each case the tension force of the cord attached to the salami must be the same in magnitude as the weight of the salami because the salami is not accelerating. Thus the scale reading is mg , where m is the mass of the salami. Its value is (11.0 kg) (9.8 m/s 2 ) = 108 N.

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10. Three vertical forces are acting on the block: the earth pulls down on the block with
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ch05 - 3 We apply Newton's second law(Eq 5-1 or...

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