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Assign3 - PHY 2130 Homework solutions Assignment 3 4.12(a...

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PHY 2130 Homework solutions Assignment 3 4.12 (a) Choose the positive y -axis in the forward direction. We resolve the forces into their components as Force x -component y -component 400 N 200 N 346 N 450 N –78.1 N 443 N Resultant 122 N x F Σ = 790 N y F Σ = The magnitude and direction of the resultant force is ( 29 ( 29 2 2 799 N x y F F = Σ + = Σ F , -1 =tan 8.77 x y F F θ Σ = ° Σ to right of y -axis. Thus, 799 N at 8.77  to the right of the forward direction = ° Σ F . (b) The acceleration is in the same direction as Σ F and is given by 799 N 3000 kg a m = = = Σ F 2 0.266  m s . 4.17 From 0 x F Σ = , 1 2 cos30.0 cos60.0 0 T T °- ° = , or ( 29 2 1 1.73 T T = . (1) Then 0 y F Σ = becomes ( 29 1 1 sin 30.0 1.73  sin 60.0 150 N 0 T T ° + °- = , which gives 1 T = 75.0 N in the right side cable . Finally, Equation (1) above gives 2 T = 130 N in the left side cable .

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4.34 First, consider the 3.00-kg rising mass. The forces on it are the tension, T , and its weight, 29.4 N. With the upward direction as positive, the second law becomes ( 29 29.4 N 3.00 kg T a - = .
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Assign3 - PHY 2130 Homework solutions Assignment 3 4.12(a...

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