429_Physics ProblemsTechnical Physics

# 429_Physics - Chapter 14 zz zb The torque is = d = rdF From the figure P14.61 = y g H y wdy = 431 H g 0 1 gwH 3 6 1 gwH 2 The total force is given

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Chapter 14 431 P14.61 The torque is ττ == zz dr d F From the figure τρ ρ =− = z ygHyw d y g w H H bg 0 3 1 6 The total force is given as 1 2 2 gwH If this were applied at a height y eff such that the torque remains unchanged, we have 1 6 1 2 32 ρρ gwH y gwH eff = L N M O Q P and yH eff = 1 3 . FIG. P14.61 P14.62 (a) The pressure on the surface of the two hemispheres is constant at all points, and the force on each element of surface area is directed along the radius of the hemispheres. The applied force along the axis must balance the force on the “effective” area, which is the projection of the actual surface onto a plane perpendicular to the x axis, AR = π 2 Therefore, FP P R 0 2 FIG. P14.62 (b) For the values given P P = = × 00 2 0 4 0 100 0 300 0 254 2 58 10 .. af m N P14.63 Looking first at the top scale and the iron block, we have: TBF g 1 += , iron where T 1 is the tension in the spring scale, B is the buoyant force, and F g , iron is the weight of the iron
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## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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