Hw 6 Solutions

# Θ tan 1 d v d h tan 1 400 m300 m and the force

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θ = tan –1 ( d v / d h ) =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 2 1 1 1 1 2 2 1 2 2 2 2 50.0 kg 9.8 m/s 1.00 m (50.0 kg)(9.8m/s )(3.00 m) = sin 3.00 m 0.800 408 N. mgx mgx T x θ + + = = (b) Equilibrium of horizontal forces requires the (rightward) horizontal hinge force be F x = T cos θ = 245 N. (c) And equilibrium of vertical forces requires the (upward) vertical hinge force be F y = mg T sin θ = 163 N. P 9 .- . (a) We note that the angle between the cable and the strut is α = θ φ = 45º – 30º = 15º. The angle between the strut and any vertical force (like the weights in the problem) is β = 90º – 45º = 45º. Denoting M = 225 kg and m = 45.0 kg, and A as the length of the boom, we compute torques about the hinge and find ( ) 2 sin sin sin sin / 2 . sin sin Mg mg Mg mg T β β β β α α + + = = A A A The unknown length A cancels out and we obtain T = 6.63 × 10 3 N.

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(b) Since the cable is at 30º from horizontal, then horizontal equilibrium of forces requires that the horizontal hinge force be 3 = cos30 = 5.74 10 N. x F T ° × (c) And vertical equilibrium of forces gives the vertical hinge force component: 3 = + + sin30 = 5.96 10 N. y F Mg mg T × D P 10 .- . (a) Computing torques about point A , we find T L Wx W L b max max sin θ = + 2 F H G I K J . We solve for the maximum distance: ( ) max max sin / 2 500sin30.0 200/ 2 = = 3.00 =1.50m. 300 b T W x L W θ °− (b) Equilibrium of horizontal forces gives max = cos = 433N. x F T θ (c) And equilibrium of vertical forces gives max = + sin = 250 N. y b F W W T θ P 11 .- Using F = GmM/r 2 , we find that the topmost mass pulls upward on the one at the origin with 1.9 × 10 8 N, and the rightmost mass pulls rightward on the one at the origin with 1.0 × 10 8 N. Thus, the ( x, y ) components of the net force, which can be converted to polar components (here we use magnitude-angle notation), are ( ) ( ) 8 8 8 net = 1.04 10 ,1.85 10 2.13 10 60.6 . F × × × ° G (a) The magnitude of the force is 2.13 × 10 8 N. (b) The direction of the force relative to the + x axis is 60.6 ° . P 12. - The acceleration due to gravity is given by a g = GM/r 2 , where M is the mass of Earth and r is the distance from Earth’s center. We substitute r = R + h , where R is the radius of Earth and h is the altitude, to obtain a g = GM / ( R + h ) 2 . We solve for
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