2_Physics Homework - Chapter 4

# 2_Physics Homework - Chapter 4 - y-component of the pull F1...

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y-component of the pull F 1 is F 1y =F 1 sin31° =(985 N)(0.5150) = x-component of the pull F 2 is F 2x = -F 2 sin32° =-(788 N)(0.5299) = Here the negative sign shows that the x-component of the pull F 2 is in the negative x direction y-component of the pull F 2 is F 2y =F 2 cos32° =-(788 N)(0.8480) = x-component of the pull F 3 is F 3x = -F 3 cos53° =-(411 N)(0.6018) = Here the negative sign shows that the x-component of the pull F 3 is in the negative x direction y-component of the pull F 3 is F 3y = -F 3 sin53° =-(411 N)(0.7986) = Here the negative sign shows that the y-component of the pull F 3 is in the negative y direction. (b) Let F be the resultant of the three pulls. Then the x- component of the F is given by And the y-component of F is given by Hence the net magnitude of the resultant force of the three pulls is: And the direction of the resultant force can be calculated as: EX 4.7 A 68.5-kg skater moving initially at 2.40 m/s on rough horizontal ice comes to rest uniformly in 3.52s due to friction from the ice. What force does the friction exert on the skater?

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