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1. A 50g weight is suspended from two cords. One cord exerts a force of 30g and makes an angle of 75° to the right of the vertical and the other exerts a force of F g and is to the left of the vertical. Find F and e. Required: Y 50g

1. A 50g weight is suspended from two cords. One cord exerts a force of 30g and makes an angle of 75° to the right of the vertical and the other exerts a force of F g and is to the left of the vertical. Find F and e. Required: Y 50g

College Physics
College Physics
1st Edition
ISBN:9781938168000
Author:Paul Peter Urone, Roger Hinrichs
Publisher:Paul Peter Urone, Roger Hinrichs
Chapter4: Dynamics: Force And Newton's Laws Of Motion
Section: Chapter Questions
Problem 21PE: Show that, as stated in the text, a force F exerted on a flexible medium at its center and...
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1. A 50g weight is suspended from two cords. One cord exerts a force of 30g and makes an angle of 75⁰ to the right of the vertical and the other exerts a force of F g and is to the left of the vertical. Find F and e. Required: F and 0 Solution: Fx=0 F = 30g cos (75°) - F cose = 0 F cose= 30 g cos (75°) F cose = 7.76g 0 = arctan (21/7.76) 0=69.72° F cos 0 = 7.76g F cos (69.72°) = 7.76g 50g Fy=0 Fy= 30g cos (75°) +F sin - W=0 W=30g sin(75°) + F sine 50g = 30g sin (75°) + F sine 50g = 29+ F sine F sin0 = 50 - 29 F sin0 = 21 F (0.35)/0.35 = 7.76/0.35 F=22.17G
Transcribed Image Text:1. A 50g weight is suspended from two cords. One cord exerts a force of 30g and makes an angle of 75⁰ to the right of the vertical and the other exerts a force of F g and is to the left of the vertical. Find F and e. Required: F and 0 Solution: Fx=0 F = 30g cos (75°) - F cose = 0 F cose= 30 g cos (75°) F cose = 7.76g 0 = arctan (21/7.76) 0=69.72° F cos 0 = 7.76g F cos (69.72°) = 7.76g 50g Fy=0 Fy= 30g cos (75°) +F sin - W=0 W=30g sin(75°) + F sine 50g = 30g sin (75°) + F sine 50g = 29+ F sine F sin0 = 50 - 29 F sin0 = 21 F (0.35)/0.35 = 7.76/0.35 F=22.17G
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