371_PartUniversity Physics Solution

371_PartUniversity Physics Solution - 11-30 11.81. Chapter...

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11-30 Chapter 11 EVALUATE: As the cable approaches the horizontal direction, its moment arm for the axis at the pivot approaches zero, so T must go to infinity in order for the torque due to the cable to continue to equal the gravity torque. 11.81. IDENTIFY: Apply the first and second conditions of equilibrium to the pole. (a) SET UP: The free-body diagram for the pole is given in Figure 11.81. n and f are the vertical and horizontal components of the force the ground exerts on the pole. x x Fm a = 0 f = The force exerted by the ground has no horizontal component. Figure 11.81 EXECUTE: 0 A τ = (7.0 m)cos (4.5 m)cos 0 Tm g θ +− = (4.5 m/7.0 m) (4.5/7.0)(5700 N) 3700 N g == = 0 y F = 0 nT m g +− = 5700 N 3700 N 2000 N nm gT = =−= The force exerted by the ground is vertical (upward) and has magnitude 2000 N. EVALUATE: We can verify that 0 z = for an axis at the cg of the pole. Tn > since T acts at a point closer to the cg and therefore has a smaller moment arm for this axis than n does. (b) In the 0 A = equation the angle divided out. All forces on the pole are vertical and their moment arms are all proportional to cos . 11.82. IDENTIFY: Apply the equilibrium conditions to the pole. The horizontal component of the tension in the wire is 22.0 N. SET UP: The free-body diagram for the pole is given in Figure 11.82. The tension in the cord equals the weight W . v F and h F are the components of the force exerted by the hinge. If either of these forces is actually in the opposite direction to what we have assumed, we will get a negative value when we solve for it. EXECUTE: (a) sin37.0 22.0 N T = ° so 36.6 N T = . 0 z = gives ( sin37.0 )(1.75 m) (1.35 m) 0 TW −= ° . (22.0 N)(1.75 m) 28.5 N 1.35 m W . (b) 0 y F = gives v cos37.0 0 FT W −− = ° and v (36.6 N)cos37.0 55.0 N 84.2 N F =+ = ° . 0 x F = gives h sin37.0 0 WT F = ° and h 28.5 N 22.0 N 6.5 N F . The magnitude of the hinge force is 22 hv 84.5 N FF F = . EVALUATE: If we consider torques about an axis at the top of the plate, we see that h F must be to the left in order for its torque to oppose the torque produced by the force W . Figure 11.82
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Equilibrium and Elasticity 11-31 11.83. IDENTIFY: Apply 0 z τ = to the slab. SET UP: The free-body diagram is given in Figure 11.83a. 3.75 m tan 1.75 m β = so 65.0 = ° . 20.0 90 βα ++= °° so 5.0 α = ° . The distance from the axis to the center of the block is 22 3.75 m 1.75 m 2.07 m ⎛⎞ += ⎜⎟ ⎝⎠ . EXECUTE: (a) (2.07 m)sin5.0 (3.75 m)sin52.0 0 wT −= . 0.061 Tw = . Each worker must exert a force of 0.012 w , where w is the weight of the slab.
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371_PartUniversity Physics Solution - 11-30 11.81. Chapter...

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