2022 Engineering Mechanics （《工程力学》全英文）(北京理工大学)1467123516 最新满分章节测试答案

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本课程起止时间为:2022-02-21到2022-06-30

【作业】01 Reductions of force systems Assignment1

小提示:本节包含奇怪的同名章节内容

1、 问题:Please see the attached pdf file.
评分规则: 【 Students should solve the problems according to the procedure taught in the lectures.
】

2、 问题:Please see the attached pdf file.

评分规则: 【 Students should solve the problems according to the procedure taught in the lectures.
】

01 Reductions of force systems Test for week 1

1、 问题:As shown in the parallelogram, all forces are at point A. In which figure FR can be considered as the resultant of F1 and F2?
选项：
A:
B:
C:
D:
答案: 【】

2、 问题:Neglect the weights of all members. In which case bar BD is not a two-force member?
选项：
A:
B:
C:
D:
答案: 【】

3、 问题:To study the whole system, which free-body diagram is correct?
选项：
A:
B:
C:
D:
答案: 【】

4、 问题:To study the whole system, which free-body diagram is correct?
选项：
A:
B:
C:
D:
答案: 【】

5、 问题:To study the whole system, which free-body diagram is correct?
选项：
A:
B:
C:
D:
答案: 【】

6、 问题:For bar AD, which free-body diagram
is correct?  
选项：
A:
B:
C:
D:
答案: 【】

7、 问题:In this equilibrium mechanism, the free-body diagram of crank OA can be drawn as
选项：
A:
B:
C:
D:
答案: 【;
;
】

8、 问题:The resultant force of two intersecting forces and is , i.e., . The magnitude of must be larger than those of and .
选项：
A:正确
B:错误
答案: 【错误】

9、 问题:As long as two forces have the same magnitude and are in opposite directions, these two forces form a couple.
选项：
A:正确
B:错误
答案: 【错误】

10、 问题:For any coplanar force system, as long as its principal vector is not equal to zero, the force system can be reduced to one single force.
选项：
A:正确
B:错误
答案: 【正确】

02 Equilibrium of force systems Test for week 2

1、 问题:All weights and frictions can be neglected. If a force P is applied at the mid-point of bar AB, in which case the bar can be in a equilibrium state?
选项：
A:
B:
C:
D:
答案: 【】

2、 问题:A rigid body is in equilibrium state under the coplanar force system. Which set of equilibrium equations are not independent of each other?
选项：
A:SX = 0, Sx = 0, SmA(F) = 0
B:SmO(F) = 0, SmA(F) = 0, SmB(F) = 0
C:SmO(F) = 0, SmC(F) = 0, SY = 0
D:SX = 0, SY = 0, SmO(F) = 0
答案: 【SmO(F) = 0, SmA(F) = 0, SmB(F) = 0】

3、 问题:A rigid body is in equilibrium state under the coplanar force system. Which set of equilibrium equations are not independent of each other?
选项：
A:SX = 0, SmO(F) = 0, SmA(F) = 0
B:SmO(F) = 0, SmA(F) = 0, SmB(F) = 0
C:SmA(F) = 0, SmC(F) = 0, SY = 0
D:SX = 0, SmA(F) = 0, SmB(F) = 0
答案: 【SmA(F) = 0, SmC(F) = 0, SY = 0】

4、 问题:Consider the equilibrium of bar AB. The correct moment equation about point A should be
选项：
A:SmA(F) = m + Psinq × L/2 + M + mA = 0
B:SmA(F) = -m – Psinq × L/2 + M = 0
C:SmA(F) = -mL – P × L/2 + M + mA = 0
D:SmA(F) = -m – Psinq × L/2 + M + mA = 0
答案: 【SmA(F) = -m – Psinq × L/2 + M + mA = 0】

5、 问题:The whole system is in equilibrium state. Which set of equilibrium equations is correct?
选项：
A:Study ADB： åmA(F) = NB×2L + YD×L = 0
B:Study the whole system： åmA(F) = – P×(L+r) + NB×2L = 0
C:Study CDE and the pully： åmD(F) = – P×r – T×(L – r) + XE×L + SBC×L×sin30° = 0
D:Study the whole system： åmA(F) = – P×(L+r) + NB×2L – T×(L – r) = 0
答案: 【Study the whole system： åmA(F) = – P×(L+r) + NB×2L – T×(L – r) = 0】

6、 问题:A planar mechanism consists of three rigid bodies and is subjected to a force system in its plane. How many independent equilibrium equations can be created at most?
选项：
A:6
B:9
C:12
D:3
答案: 【9】

7、 问题:The whole system is in equilibrium state. Assume the member force in bar i (i=1,2,…5) is Si. Which set of equilibrium
equations is correct?
选项：
A:Study AB： åmA(F) = -q×L×L/2 + S2 ×L/2 – P×3L/4 = 0
B:Study ABCDE： åmA(F) = -q×2L×L + S2 ×L/2 + S3×3L/2 – P×3L/4 + YB×L + NC×2L = 0
C:Study BC： åmB(F) = -q×L×L/2 – S3 ×L/2 – S5×L/2 + NC×L = 0
D:Study ABCDE： åmA(F) = -q×2L×L – P×3L/4 + NC×2L = 0
答案: 【Study ABCDE： åmA(F) = -q×2L×L – P×3L/4 + NC×2L = 0】

8、 问题:Consider the equilibrium of the whole system. Which sets of equilibrium equations are correct?
选项：
A:åmO(F) = – m – XA×L×sinj + YA×L×cosj – Q×2L×cosj – P× L×sinj + NB×3L×cosj = 0
B:åmO(F) = – m – Q×2L×cosj – P× L×sinj + NB×3L×cosj = 0
C:åmB(F) = – m – XA×2L×sinj – YA×2L×cosj + Q×L×cosj = 0
D:åmB(F) = – m – XO× L×sinj – YO×3L×cosj+ Q×L×cosj = 0
答案: 【åmO(F) = – m – Q×2L×cosj – P× L×sinj + NB×3L×cosj = 0;
åmB(F) = – m – XO× L×sinj – YO×3L×cosj+ Q×L×cosj = 0】

9、 问题:For a free-body diagram in one plane, which sets of equilibrium equations could be independent of each other?
选项：
A:three force equations
B:two force equations and one moment equation
C:one force equation and two moment equations
D:three moment equations
答案: 【two force equations and one moment equation;
one force equation and two moment equations;
three moment equations】

10、 问题:An action and its reaction have the same magnitude and are in opposite directions and collinear, so they form an equilibrium force system.
选项：
A:正确
B:错误
答案: 【错误】

【作业】02 Equilibrium of force systems Assignment2

1、 问题:Please see the attached pdf file.
评分规则: 【 Students should solve the problems according to the procedure taught in
the lectures.
】

【作业】05 Composite motion of a point Assignment3

小提示:本节包含奇怪的同名章节内容

1、 问题:Please see the attached pdf file.