Quiz3[UnknownDate] - (1 If the arm(Fg weighs 41.5 N.The...

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(1) If the arm (Fg) weighs 41.5 N .The force of gravity acting on the arms acts through point A. Determine the magnitudes of the tension force F t in the deltoid muscle and the force F s exerted by the shoulder on the homeruns (upper-arm bone) to hold the arm in the position shown Solution Since the arm is static then : Σ F = 0 and Στ = 0 Selecting point O as the pivot. Then: Ft x 0.08 sin12 – Fg x 0.29 = 0, then Ft = 41.5 x 0.29 / 0.08 sin12 = 723.6 N Σ Fy = Ft sin12 – Fg – Fs sin θ = 0, then Fs sin θ = 723.6 sin12 – 41.5 = 108.9 Σ Fx = Fs cos θ – Ft cos12 = 0, then Fs cos θ = 723.6 cos12 = 707.7 g1832g1871 g1871g1861g1866g2016 g1832g1871 g1855g1867g1871g2016 g3404 108.9 707.7 g34040.154 Then tan θ = 0.154 so θ = tan -1 (0.145) = 8.74 o And Fs = 707.7/cos8.74 = 718 N (2) The horizontal arm is composed of three parts: the upper arm (W 1 =17 N), the lower arm (W 2 =11 N), and the hand (W 3 = 4.2 N). Find: (i) The magnitude of the force needed to stop the arm falling down. (ii) If the force is pointing straight upwards, what would be its distance from the shoulder joint? Solution The arm is static then Σ F = 0 and Στ = 0 F + W1 + W2 + W3 = 0 F – 17 – 11 – 4.2 = 0 then F = 32.2 N F l + W 1 l 1 + W 2 l 2 + W 3 l 3 = 0, then 32.2 l = 17x0.13 + 11x0.38 + 4.2x0.61 and l = 0.278 m
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(3) A child of mass 35 kg slides down a smooth slide as shown. Calculate:
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