356_PartUniversity Physics Solution

356_PartUniversity Physics Solution - Equilibrium and...

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Equilibrium and Elasticity 11-15 EVALUATE: We assumed that the wing force was upward and the tail force was downward. When we solved for these forces we obtained positive values for them, which confirms that they do have these directions. Note that the rear stabilizer provides a downward force. It does not hold up the tail of the aircraft, but serves to counter the torque produced by the wing. Thus balance, along with weight, is a crucial factor in airplane loading. Figure 11.43 11.44. IDENTIFY: Apply the first and second conditions of equilibrium to the truck. SET UP: The weight on the front wheels is f n , the normal force exerted by the ground on the front wheels. The weight on the rear wheels is r n , the normal force exerted by the ground on the rear wheels. When the front wheels come off the ground, f 0 n . The free-body diagram for the truck without the box is given in Figure 11.44a and with the box in Figure 11.44b. The center of gravity of the truck, without the box, is a distance x from the rear wheels. EXECUTE: 0 y F = in Fig.11.44a gives rf 8820 N 10,780 N 19,600 N wn n =+= + = 0 τ = in Fig.11.44a, with the axis at the rear wheels and counterclockwise torques positive, gives f (3.00 m) 0 nw x −= and f (3.00 m) 10,780 N (3.00 m) 1.65 m 19,600 N n x w ⎛⎞ == = ⎜⎟ ⎝⎠ . (a) 0 = in Fig.11.44b, with the axis at the rear wheels and counterclockwise torques positive, gives box f (1.00 m) (3.00 m) (1.65 m) 0 wn w +− = . f (3600 N)(1.00 m) (19,600 N)(1.65 m) 9,580 N 3.00 m n −+ 0 y F = gives r f box nnw w += + and r 3600 N 19,600 N 9580 N 13,620 N n =+ . There is 9,580 N on the front wheels and 13,620 N on the rear wheels. (b) f 0 n . 0 = gives box (1.00 m) (1.65 m) 0 ww = and 4 box 1.65 3.23 10 N × . EVALUATE: Placing the box on the tailgate in part (b) reduces the normal force exerted at the front wheels. Figure 11.44a, b 11.45. IDENTIFY: In each case, to achieve balance the center of gravity of the system must be at the fulcrum. Use Eq.(11.3) to locate cm x , with i m replaced by i w . SET UP: Let the origin be at the left-hand end of the rod and take the x + axis to lie along the rod. Let 1 255 N w = (the rod) so 1 1.00 m x = , let 2 225 N w = so 2 2.00 m x = and let 3 wW = . In part (a) 3 0.500 m x = and in part (b) 3 0.750 m x = . EXECUTE: (a) cm 1.25 m x = . 11 2 2 33 cm 123 wx x www ++ = gives 12 c m1 2 3 3c m () wwx w xw x w xx = and (480 N)(1.25 m) (255 N)(1.00 m) (225 N)(2.00 m) 140 N 0.500 m 1.25 m W −− . (b) Now 3 140 N and 3 0.750 m x = . cm (255 N)(1.00 m) (225 N)(2.00 m) (140 N)(0.750 m) 1.31 m 255 N 225 N 140 N x . W must be moved 1.31 m 1.25 m 6 cm to the right. EVALUATE: Moving W to the right means cm x for the system moves to the right.
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11-16 Chapter 11 11.46. IDENTIFY: The center of gravity of the object must have the same x coordinate as the hook. Use Eq.(11.3) for cm x . The mass of a segment is proportional to its length. Define α to be the mass per unit length, so ii ml = , where i l is the length of a piece that has mass i m .
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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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356_PartUniversity Physics Solution - Equilibrium and...

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