366_PartUniversity Physics Solution

366_PartUniversity Physics Solution - Equilibrium and...

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Equilibrium and Elasticity 11-25 EVALUATE: As expected, the legs closest to the 1500 N weight exert a greater force on the floor. Figure 11.70a, b 11.71. IDENTIFY: Apply 0 z τ = first to the roof and then to one wall. (a) SET UP: Consider the forces on the roof; see Figure 11.71a. V and H are the vertical and horizontal forces each wall exerts on the roof. 20,000 N w = is the total weight of the roof. 2 Vw = so /2 = Figure 11.71a Apply 0 z = to one half of the roof, with the axis along the line where the two halves join. Let each half have length L . EXECUTE: ( / 2)( /2)(cos35.0 ) sin35.0 cos35 0 wL H L V L °+ °− °= L divides out, and use = 1 4 sin35.0 cos35.0 Hw ° 7140 N 4tan35.0 w H == ° EVALUATE: By Newton&s 3rd law, the roof exerts a horizontal, outward force on the wall. For torque about an axis at the lower end of the wall, at the ground, this force has a larger moment arm and hence larger torque the taller the walls. (b) SET UP: The force diagram for one wall is given in Figure 11.71b. Consider the torques on this wall. Figure 11.71b H is the horizontal force exerted by the roof, as considered in part (a). B is the horizontal force exerted by the buttress. Now the angle is 40 , ° so 5959 N 4tan40 w H ° EXECUTE: 0, z = axis at the ground (40 m) (30 m) 0 HB −= and 7900 N. B = EVALUATE: The horizontal force exerted by the roof is larger as the roof becomes more horizontal, since for torques applied to the roof the moment arm for H decreases. The force B required from the buttress is less the higher up on the wall this force is applied. 11.72. IDENTIFY: Apply 0 z = to the wheel. SET UP: Take torques about the upper corner of the curb. EXECUTE: The force F ! acts at a perpendicular distance R h and the weight acts at a perpendicular distance () 2 22 2. R Rh R hh −− = Setting the torques equal for the minimum necessary force, 2 2 . R Fm g =
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11-26 Chapter 11 (b) The torque due to gravity is the same, but the force F ! acts at a perpendicular distance 2 , R h so the minimum force is () 2 / ( 2 ) . mg Rh hv R h −− EVALUATE: (c) Less force is required when the force is applied at the top of the wheel, since in this case F # has a larger moment arm. 11.73. IDENTIFY: Apply the first and second conditions of equilibrium to the gate. SET UP: The free-body diagram for the gate is given in Figure 11.73. Figure 11.73 Use coordinates with the origin at B. Let A H ! and B H ! be the forces exerted by the hinges at A and B. The problem states that A H ! has no horizontal component. Replace the tension T ! by its horizontal and vertical components.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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366_PartUniversity Physics Solution - Equilibrium and...

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