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HW9-phy2048-Spring-2014-solution

# 5 1010 nm2 strain 0002 b since the linear range of the

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Unformatted text preview: angle where the cable comes into contact (also at x2) is θ = tan–1(dv/dh) = tan–1(4.00 m/3.00 m) and the force exerted there is the tension T. Computing torques about the hinge, we find 2 1 1 1 (1.00 m ) + 1 (50.0 kg) (9.8m/s 2 ) (3.00 m) 2 ( 50.0 kg ) 9.8 m/s 2 2 mgx1 + 2 mgx2 T= = x sinθ (3.00 m ) ( 0.800 ) ( ) 2 = 408 N. (b) Equilibrium of horizontal forces requires that the horizontal hinge force be Fx = T cos θ = 245 N. 2 PHY2048 Spring 2014 HW9 (c) The direction of the horizontal force is rightward. (d) Equilibrium of vertical forces requires that the vertical hinge force be Fy = mg – T sin θ = 163 N. (e) The direction of the vertical force is upward. 12P44. (a) The Young’s modulus is given by stress 150 ×106 N/m2 E= = slope of the stress-strain curve = = 7.5 ×1010 N/m2 . strain 0.002 (b) Since the linear range of the curve extends...
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