CW5 - boland (kbb544) – CW5 – Mackie – (10611) 1 This...

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Unformatted text preview: boland (kbb544) – CW5 – Mackie – (10611) 1 This print-out should have 4 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points Two forces, 434 N at 16 ◦ and 289 N at 33 ◦ are applied to a car in an effort to accelerate it. 3100 kg 4 3 4 N 16 ◦ 2 8 9 N 3 3 ◦ a) Find the resultant of these two forces. Correct answer: 660 . 644 N. Explanation: Given : m = 3100 kg , F 1 = 434 N , θ 1 = 16 ◦ , F 2 = 289 N , and θ 2 =- 33 ◦ . Consider the side forces: F 1 ,y = F 1 sin θ 1 F 2 ,y = F 2 sin θ 2 F y,net = (434 N) sin16 ◦ + (289 N) sin(- 33 ◦ ) =- 37 . 7741 N Consider the forward forces: F 1 ,x = F 1 cos θ 1 F 2 ,x = F 2 cos θ 2 F x,net = (434 N) cos16 ◦ + (289 N) cos(- 33 ◦ ) = 659 . 563 N Thus the net force is F net = radicalBig (659 . 563 N) 2 + (- 37 . 7741 N) 2 = 660 . 644 N ....
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This note was uploaded on 02/20/2011 for the course PHYS 1062 taught by Professor Mackie during the Spring '11 term at Temple.

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CW5 - boland (kbb544) – CW5 – Mackie – (10611) 1 This...

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