M11 YOUN7066 13 ISM c11 tủ tài liệu training
➤ Gửi thông báo lỗi ⚠️ Báo cáo tài liệu vi phạmNội dung chi tiết: M11 YOUN7066 13 ISM c11 tủ tài liệu training
M11 YOUN7066 13 ISM c11 tủ tài liệu training
11Equilibrium AM) Elasticity11.1.Identify: Use Eq. (11.3) to calculate x^. The center of gravity of the bat IS at Its center and it can be treated as M11 YOUN7066 13 ISM c11 tủ tài liệu training a point mass at that point.Set Up: Use coordinates with the origin at the left end of the bar and the +.V axis along the bar.wtj = 0.120 kg. 1/h = 0.055 kg. in3 = 0.110 kg.Exectte: ,v.„ = ZỈỈ£l±£ỉ2ĨL±» = (0120 kg)(0250 m) + 0 + (0.110 kg)(0300 m) = 0 298 m The;»J + »tj + IMJ0.120 kg + 0.055 kg 4- 0 M11 YOUN7066 13 ISM c11 tủ tài liệu training.110 kgfulcrum 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 lefM11 YOUN7066 13 ISM c11 tủ tài liệu training
t-hand end. So the center of gravity is to the right of the ceutei of the bat11.2.Identify: Use Eq. (11.3) to calculate x^ of the composite object.Set11Equilibrium AM) Elasticity11.1.Identify: Use Eq. (11.3) to calculate x^. The center of gravity of the bat IS at Its center and it can be treated as M11 YOUN7066 13 ISM c11 tủ tài liệu training0 cm. = 5.00 g and =1.50 g. Xj = 0... W'X' ..+ _i 5.00 g +1.50 g '1. _..Execute: x,_ = —X =1 — - X = --------------£ (-2.20 cm) = -9.53 cm.»>i + nhk w2 )\1.50 g )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 somewh M11 YOUN7066 13 ISM c11 tủ tài liệu trainingere betw een the added mass and the original center of gravity.11.3.Identify: Treat the rod and clamp as point masses. The center of gravity of the roM11 YOUN7066 13 ISM c11 tủ tài liệu training
d IS at Its midpoint, and we know the location of the center of gravity of the rod-clamp system._____;w.X| + whXjSETUP: xcm=»»J + J»2_______ ,(1.80 kg11Equilibrium AM) Elasticity11.1.Identify: Use Eq. (11.3) to calculate x^. The center of gravity of the bat IS at Its center and it can be treated as M11 YOUN7066 13 ISM c11 tủ tài liệu traininghe right of the center of gravity of the system, so the centei of gravity of the system lies between that of the rod and the damp, which IS reasonable.11.4.Identify: Apply the first and second conditions for equilibrium to tile nap door.Set Up: For z r. = 0 take the axis at the hinge. Then the torqu M11 YOUN7066 13 ISM c11 tủ tài liệu traininge 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 apM11 YOUN7066 13 ISM c11 tủ tài liệu training
plied force must have the same magnitude as the weight of the door, or 300 N. In this case the binge exerts no force.(b) With respect to the hinges, t11Equilibrium AM) Elasticity11.1.Identify: Use Eq. (11.3) to calculate x^. The center of gravity of the bat IS at Its center and it can be treated as M11 YOUN7066 13 ISM c11 tủ tài liệu training 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.Illc Copyright 2012 Pearson Education. Inc All rights reserved This material is protected under all copyright laws as tliey currently exist.No pcction of this materul may b M11 YOUN7066 13 ISM c11 tủ tài liệu traininge reproduced, in any form or by any means, without permission in writing from the publisher.11-2 Chapter 1111.5.Identify: Apply Zr_-=0 to the ladder.SM11 YOUN7066 13 ISM c11 tủ tài liệu training
et Up: Take the axis to be at point .4 The fiee-body diagram for the ladder is given in Figure 11 5 The torque due to F must balance the torque due to11Equilibrium AM) Elasticity11.1.Identify: Use Eq. (11.3) to calculate x^. The center of gravity of the bat IS at Its center and it can be treated as 11Equilibrium AM) Elasticity11.1.Identify: Use Eq. (11.3) to calculate x^. The center of gravity of the bat IS at Its center and it can be treated asGọi ngay
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