YF_ISM_11

YF_ISM_11 - EQUILIBRIUM AND ELASTICITY 11 11.1. IDENTIFY:...

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11-1 E QUILIBRIUM AND E LASTICITY 11.1. IDENTIFY: Use Eq.(11.3) to calculate cm x . The center of gravity of the bar is at its center and it can be treated as a point mass at that point. SET UP: Use coordinates with the origin at the left end of the bar and the x + axis along the bar. 1 2.40 kg, m = 2 1.10 kg, m = 3 2.20 kg. m = EXECUTE: 11 2 2 33 cm 123 (2.40 kg)(0.250 m) 0 (2.20 kg)(0.500 m) 0.298 m 2.40 kg 1.10 kg 2.20 kg mx x mmm ++ + + == = + + . The fulcrum should be placed 29.8 cm to the right of the left-hand end. EVALUATE: The mass at the right-hand end is greater than the mass at the left-hand end. So the center of gravity is to the right of the center of the bar. 11.2. IDENTIFY: Use Eq.(11.3) to calculate cm x of the composite object. SET UP: Use coordinates where the origin is at the original center of gravity of the object and x + is to the right. With the 1.50 g mass added, cm 2.20 cm x =− , 1 5.00 g m = and 2 1.50 g m = . 1 0 x = . EXECUTE: 22 cm 12 x mm = + . 2c m 2 5.00 g 1.50 g (2 . 2 0 cm ) 9 . 5 3 1.50 g xx m ⎛⎞ = ⎜⎟ ⎝⎠ . The additional mass should be attached 9.53 cm to the left of the original center of gravity. EVALUATE: The new center of gravity is somewhere between the added mass and the original center of gravity. 11.3. IDENTIFY: The center of gravity of the combined object must be at the fulcrum. Use Eq.(11.3) to calculate cm x SET UP: The center of gravity of the sand is at the middle of the box. Use coordinates with the origin at the fulcrum and x + to the right. Let 1 25.0 kg m = , so 1 0.500 m x = . Let 2s a n d = , so 2 0.625 m x . cm 0 x = . EXECUTE: cm 0 x + + and 1 21 2 0.500 m (25.0 kg) 20.0 kg 0.625 m x x = . EVALUATE: The mass of sand required is less than the mass of the plank since the center of the box is farther from the fulcrum than the center of gravity of the plank is. 11.4. IDENTIFY: Apply the first and second conditions for equilibrium to the trap door. SET UP: For 0 z τ = take the axis at the hinge. Then the torque due to the applied force must balance the torque due to the weight of the door. EXECUTE: (a) The force is applied at the center of gravity, so the applied force must have the same magnitude as the weight of the door, or 300 N. In this case the hinge exerts no force. (b) With respect to the hinges, the moment arm of the applied force is twice the distance to the center of mass, so the force has half the magnitude of the weight, or 150 N . The hinges supply an upward force of 300 N 150 N 150 N. −= EVALUATE: Less force must be applied when it is applied farther from the hinges. 11.5. IDENTIFY: Apply 0 z = to the ladder. SET UP: Take the axis to be at point A . The free-body diagram for the ladder is given in Figure 11.5. The torque due to F must balance the torque due to the weight of the ladder. EXECUTE: (8.0 m)sin40 (2800 N)(10.0 m), so 5.45 kN FF °= = . 11
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11-2 Chapter 11 EVALUATE: The force required is greater than the weight of the ladder, because the moment arm for F is less than the moment arm for w .
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This note was uploaded on 03/14/2012 for the course MAE 162D taught by Professor Shaefer during the Spring '11 term at UCLA.

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YF_ISM_11 - EQUILIBRIUM AND ELASTICITY 11 11.1. IDENTIFY:...

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