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起止时间:2021-02-26到2021-06-30
更新状态:已完结
作业01 Reductions of force systems Assignment1
1、 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 Tests for week 1
1、 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?
A:
B:
C:
D:
答案:
2、 Neglect the weights of all members. In which case bar BD is not a two-force member?
A:
B:
C:
D:
答案:
3、 To study the whole system, which free-body diagram is correct? 
A:
B:
C:
D:
答案:
4、 To study the whole system, which free-body diagram is correct?
A:
B:
C:
D:
答案:
5、 To study the whole system, which free-body diagram is correct? 
A:
B:
C:
D:
答案:
6、 For bar AD, which free-body diagram
is correct? 
A:
B:
C:
D:
答案:
7、 In this equilibrium mechanism, the free-body diagram of crank OA can be drawn as
A:
B:
C:
D:
答案: ;
;
8、 The resultant force of two intersecting forces 
and 
is 
, i.e., 
. The magnitude of must be larger than those of and .
A:正确
B:错误
答案: 错误
分析：FR is the vector summation of F1 and F2, so it’s magnitude could be larger or smaller than those of F1 and F2.
9、 As long as two forces have the same magnitude and are in opposite directions, these two forces form a couple.
A:正确
B:错误
答案: 错误
分析：They must be separated by a perpendicular distance.
10、 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.
A:正确
B:错误
答案: 正确
作业02 Equilibrium of force systems Assignment2
1、 Please see the attached pdf file.
评分规则: Students should solve the problems according to the procedure taught in
the lectures.
02 Equilibrium of force systems Tests for week 2
1、 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?
A:
B:
C:
D:
答案:
2、 A rigid body is in equilibrium state under the coplanar force system. Which set of equilibrium equations are not independent of each other? 
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、 A rigid body is in equilibrium state under the coplanar force system. Which set of equilibrium equations are not independent of each other?
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、 Consider the equilibrium of bar AB. The correct moment equation about point A should be 
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、 The whole system is in equilibrium state. Which set of equilibrium equations is correct? 
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、 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?
A:6
B:9
C:12
D:3
答案: 9
7、 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?
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、 Consider the equilibrium of the whole system. Which sets of equilibrium equations are correct? 
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、 For a free-body diagram in one plane, which sets of equilibrium equations could be independent of each other?
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、 An action and its reaction have the same magnitude and are in opposite directions and collinear, so they form an equilibrium force system.
A:正确
B:错误
答案: 错误
分析：Action and reaction are internal forces between different bodies. An equilibrium force system must be external forces for one body or a body system.
作业05 Composite motion of a point Assignment3
1、 Please see the attached pdf file.
评分规则: Students should solve the problems according to the procedure taught in the lectures.
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