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(b) Solving the above equation algebraically for F , with W denoting the weight, we obtain (0.42)(180 N) 76 N . cos sin cos (0.42) sin cos (0.42) sin s s W F μ θ θθ == = ++ + (c) We minimize the above expression for F by working through the condition: 2 (sin cos ) 0 (cos sin ) ss s W dF d − + which leads to the result = tan –1 s = 23°. (d) Plugging = 23° into the above result for F , with s = 0.42 and W = 180 N, yields 70 N F = . 32. We use the familiar horizontal and vertical axes for x and y directions, with rightward and upward positive, respectively. The rope is assumed massless so that the force exerted by the child G F is identical to the tension uniformly through the rope. The x and y components of
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