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A3 2010 PHYS115-handoutolutions

# A3 2010 PHYS115-handoutolutions - F10-PHYS115-UAE 45...

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F10-PHYS115-UAE Assignment 3 Solutions 1/4 45. INTERPRET This problem deals with interaction between different pairs of objects. The key concepts involved here are Newton’s second and third laws. DEVELOP Let the three masses be denoted, from left to right, as 12 3 ,, a n d, mm m respectively. We take the right direction to be x . Assume that the table surface is horizontal and frictionless so that the only horizontal forces are the applied force and the contact forces between the blocks. For example, 12 F denotes the force exerted by block 1 on block 2. Since the blocks are in contact, they all have the same acceleration a , to the right. Newton’s second law can be applied to each block separately: app 21 1 12 32 2 23 3 F Fm a F a F ma   EVALUATE Adding all three equations and using Newton’s third law 12 21 (0 , FF etc.), one finds app 2 123 12 N 2 m/s (to the right) 1kg 2kg 3kg F a mmm     Thus, the force block 2 exerts on block 3 is 2 23 3 (3 kg)(2.0 m/s ) 6.0 N a (to the right). ASSESS If we draw a free-body diagram for 3 , m we will see that 23 F is the only force acting on it (ignoring friction). Thus, the force that accelerates 3 m comes entirely from 23 . F 47. INTERPRET The objects of interest are the airplane and the two gliders, connected by ropes. The problem involves applying Newton’s second law to find horizontal thrust of the plane’s propeller, the tension forces in each rope, and the net force on the first glider. DEVELOP Let the masses of the airplane and the two gliders be denoted, from right to left, as 3 a n d m , respectively (see figure). We take the right direction to be x . Assume that the runway surface is horizontal and frictionless so that the only horizontal forces are the thrust and the tensions in the ropes. We may then write the horizontal component (positive in direction of the acceleration) of the equations of motion (Newton’s second law) for the three planes (all assumed to have the same acceleration) as follows: th 1 1 12 2 23 (airplane) (first glider) (second glider) FTm a TT m a Tm a  Note: The tension has the same magnitude at every point in a rope of negligible mass. EVALUATE (a) Add all the equations of motion (the tensions cancel in pairs due to Newton’s third law), the thrust of the airplane is th 1 2 3 ( ) (2200kg 310kg 260kg)(1.9 m/s ) 5.26 10 N m m a  (b) The tension in the first rope is 1t h1 ( ) (310 kg 260kg)(1.9 m/s ) 1.08 10 N TFm amm a  (c) The tension in the second rope is 2 (260 kg)(1.9 m/s ) 494 N. a (d) The net force on the first glider is 2 net 2 (310 kg)(1.9 m/s ) 589 N. a

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F10-PHYS115-UAE Assignment 3 Solutions 2/4 ASSESS
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A3 2010 PHYS115-handoutolutions - F10-PHYS115-UAE 45...

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