HW_4-Solution (1).pptx

HW_4-Solution (1).pptx - a 60 cm With the hand placed in...

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y x z 60 cm COM Z2=100 cm Z1=50 cm z y R2 W R1 Let R1 be reaction at hand; R2 be reaction at leg. Using force equilibrium equations in Y-Z plane Fy = 0 ; no forces in z-directions 2R2 + R1 = W M about X-axis at hand = 0 Z1 W – (Z1+Z2) 2 R2 = 0 50 x 700 – 150 x 2 R2 = 0 R2 = 35,000 / 300 = 116.67 N R1 = 700 – 2 x 116.67 = 466.67 N a) With the hand placed in this position determine all ground reactions for the hand and the two feet (YZ plane)
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b) If the maximum z-directed moment the hand can support without the bodybuilder falling over is 315 Nm what is the maximum distance in the x-direction from the center of mass the hand can be placed? (XY plane) R2 R2 60 cm R1 Hand Legs Xm R1 reaction at hand at Xm from COM in XY plane R2 is reaction at leg. Let Mg =315 Nm = 31,500 Ncm Using force equilibrium equations in Y-Z plane Fy = 0 ; no forces in z-directions 2R2 + R1 = W M about Z-axis at hand = 0 Mg – (Xm * W) + (Xm+ 30) R2 + (Xm – 30) R2= 0 31,500 – (700Xm) + 2* 116.67 Xm = 0 Xm = 31,500 / 466.67 = 67.5 cm X y
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c) If we assume that the triceps is the only muscle supporting the forearm then draw FBDs of the lower arm/hand and determine the magnitude of the triceps force, and the magnitude and
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