Problem_Set_3_Solutions

Problem_Set_3_Solutions - 4. We note that m a = (16 N) i +...

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4. 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 = F 1 F 2 =(–34 N) i ^ (12 N) j ^ .
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32. We resolve this horizontal force into appropriate components. (a) Newton’s second law applied to the x -axis produces Fm g m a cos sin . θ −= For a = 0, this yields F = 566 N. (b) Applying Newton’s second law to the y axis (where there is no acceleration), we have sin cos 0 N FF m g θθ −− = which yields the normal force F N = 1.13 × 10 3 N.
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mg F w sin . θ −= 0 This yields F w = 68 N (uphill). (b) Given our coordinate choice, we have a =| a |= 1.0 m/s 2 . Newton’s second law mg F ma w now leads to F w = 28 N (uphill). (c) Continuing with the forces as shown in our figure, the equation mg F ma w will lead to F w = – 12 N when | a | = 2.0 m/s 2 . This simply tells us that the wind is opposite to the direction shown in our sketch; in other words, 12 N w F = G downhill . 36. We label the 40 kg skier “ m ” which is represented as a block in the figure shown. The force of the wind is denoted G F w and might be either “uphill” or “downhill” (it is shown uphill in our sketch). The incline angle is 10°. The x direction is downhill. (a) Constant velocity implies zero acceleration; thus, application of Newton’s second law along the x axis leads to
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50. First, we consider all the penguins (1 through 4, counting left to right) as one system, to which we apply Newton’s second law: () ( ) 4 1234 2 222N 12kg 15kg 20kg .
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This note was uploaded on 11/16/2009 for the course PHYSICS 11a taught by Professor Silvera during the Fall '07 term at Harvard.

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Problem_Set_3_Solutions - 4. We note that m a = (16 N) i +...

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