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Unformatted text preview: 5.1: a) The tension in the rope must be equal to each suspended weight, 25.0 N. b) If the mass of the light pulley may be neglected, the net force on the pulley is the vector sum of the tension in the chain and the tensions in the two parts of the rope; for the pulley to be in equilibrium, the tension in the chain is twice the tension in the rope, or 50.0 N. 5.2: In all cases, each string is supporting a weight w against gravity, and the tension in each string is w . Two forces act on each mass : w down and ) ( w T = up. 5.3: a) The two sides of the rope each exert a force with vertical component T θ sin , and the sum of these components is the hero’s weight. Solving for the tension T , . N 10 54 . 2 0.0 1 sin 2 ) s m (9.80 kg) . 90 ( sin 2 3 2 × = ° = θ = w T b) When the tension is at its maximum value, solving the above equation for the angle θ gives . 01 . 1 N) 10 50 . 2 ( 2 s m (9.80 kg) (90.0 arcsin 2 arcsin 4 2 ° = × = = T w θ 5.4: The vertical component of the force due to the tension in each wire must be half of the weight, and this in turn is the tension multiplied by the cosine of the angle each wire makes with the vertical, so if the weight is . 48 arccos and cos , 3 2 4 3 2 ° = = = θ θ w w w 5.5: With the positive ydirection up and the positive xdirection to the right, the free body diagram of Fig. 5.4(b) will have the forces labeled n and T resolved into x and y components, and setting the net force equal to zero, . sin cos sin cos = + = = = w T n F n T F y x α α α α Solving the first for α T n cot = and substituting into the second gives w T T T T = = + = + α α α α α α α α sin sin sin sin cos sin sin cos 2 2 2 and so , cos cot sin cot α = α α = α = w w T n as in Example 5.4. 5.6: N. 10 4.10 17.5 sin ) s m (9.80 kg) 1390 ( sin sin 3 2 × = ° = = α mg α w 5.7: a) N. 10 23 . 5 cos or , cos 4 40 cos ) s m kg)(9.8 4090 ( 2 × = = = = ° θ W T W θ T B B b) N. 10 36 . 3 40 sin N) 10 23 . 5 ( sin 4 4 × = ° × = = θ T T B A 5.8: a) . 45 cos 30 cos and , 45 sin 30 sin , = ° ° = = ° + ° = B A C B A C T T w T T T w T Since , 45 cos 45 sin ° = ° adding the last two equations gives , ) 30 sin 30 (cos w T A = ° + ° and so . 732 . 366 . 1 w T w A = = Then, . 897 . 45 cos 30 cos w T T A B = = ° ° b) Similar to part (a), , 45 sin 60 cos , w T T w T B A C = ° + ° = and . 45 cos 60 sin = ° ° B A T T Again adding the last two, , 73 . 2 ) 60 cos 60 (sin w T w A = = ° ° and . 35 . 3 45 cos 60 sin w T T B B = = ° ° 5.9: The resistive force is N. 523 m) 6000 m 200 )( s m kg)(9.80 1600 ( sin 2 = = α w . 5.10: The magnitude of the force must be equal to the component of the weight along the incline, or N. 337 . 11 sin ) s m kg)(9.80 180 ( sin 2 = ° = θ W 5.11: a) , sin N, 60 W θ T W = = so , sin45 N) 60 ( ° = T or N. 85 = T b) N. 60 45 cos N 85 , cos 2 1 2 1 = ° = = = = F F θ T F F 5.12: If the rope makes an angle...
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This homework help was uploaded on 04/08/2008 for the course PHY 205 taught by Professor Nearing during the Fall '07 term at University of Miami.
 Fall '07
 Nearing
 Physics, Force, Mass, Light

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