# D the implication is that what as needed was 22300 kg

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(d) The implication is that what as needed was 22300 kg – 6172 kg = 16128 kg, so the request should have been for (16128 kg)/(0.8034 kg/L) = 20075 L. (e) The percentage of the required fuel was 7682 L (on board) + 4917 L (added) (22300 kg required) /(0.8034 kg/L) = 45%.

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70. We are only concerned with horizontal forces in this problem (gravity plays no direct role). Without loss of generality, we take one of the forces along the + x direction and the other at 80 ° (measured counterclockwise from the x axis). This calculation is efficiently implemented on a vector capable calculator in polar mode, as follows (using magnitude- angle notation, with angles understood to be in degrees): F net = (20 0) + (35 80) = (43 53) ¡ | F net | = 43 N . Therefore, the mass is m = (43 N)/(20 m/s 2 ) = 2.2 kg.
a F F F m = = = 1 2 3 2 50 083 N 30N 10N 12 kg m / s . . (c) In this case, the forces G G F F 2 3 and are collectively strong enough to have y components (one positive and one negative) which cancel each other and still have enough x contributions (in the – x direction) to cancel G F 1 . Since G G F F 2 3 = , we see that the angle above the – x axis to one of them should equal the angle below the – x axis to the other one (we denote this angle θ ). We require ( ) ( ) 2 3 50 N 30N cos 30N cos x x F F θ θ = + = − which leads to θ = F H G I K J = ° cos . 1 50 34 N 60N 71. The goal is to arrive at the least magnitude of G F net , and as long as the magnitudes of G F 2 and G F 3 are (in total) less than or equal to G F 1 then we should orient them opposite to the direction of G F 1 (which is the + x direction). (a) We orient both G G F F 2 3 and in the – x direction. Then, the magnitude of the net force is 50 – 30 – 20 = 0, resulting in zero acceleration for the tire. (b) We again orient G G F F 2 3 and in the negative x direction. We obtain an acceleration along the + x axis with magnitude

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72. (a) A small segment of the rope has mass and is pulled down by the gravitational force of the Earth. Equilibrium is reached because neighboring portions of the rope pull up sufficiently on it. Since tension is a force along the rope, at least one of the neighboring portions must slope up away from the segment we are considering. Then, the tension has an upward component which means the rope sags. (b) The only force acting with a horizontal component is the applied force G F . Treating the block and rope as a single object, we write Newton’s second law for it: F = ( M + m ) a , where a is the acceleration and the positive direction is taken to be to the right. The acceleration is given by a = F /( M + m ). (c) The force of the rope F r is the only force with a horizontal component acting on the block. Then Newton’s second law for the block gives F Ma MF M m r = = + where the expression found above for a has been used. (d) Treating the block and half the rope as a single object, with mass 1 2 M m + , where the horizontal force on it is the tension T m at the midpoint of the rope, we use Newton’s second law: ( ) ( ) ( ) ( ) / 2 2 1 .

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