工程力学课后习题答案第五章 空间任意力系
第五章 空间任意力系
5.1解:F x =F cos 45 sin 60 =1.22K N F y =F c o s 45c o s =60
F z =F sin 45=1.4K N M M
y
K 0N . 7
x
=F z 60m m =84. K 8N 5⋅m m
z
=F z 50m m =70.71K N ⋅m m M =F x 60m m +F 50m m =108. 84K ⋅N m m y
β5.2 解:F x =F 2sin α-F 1cos βsin α F y =-F 1c o s
F z =F 1sin β+F 2cos αM M
y
c o αs
x
=F z a =aF 1sin β+aF 2cos α
=aF 1sin β M
z
=F y a -F x a =-aF 1cos βcos α-aF 2sin α-aF 1cos βsin α
5.3解:两力F 、F ′能形成力矩M
1
M 1=Fa =⋅m M 1x =M 1cos 45M 1y =0 M 1z =M 1sin 45
M x =M 1c o s 45=
5K 0N ⋅ m M
z
=M 1z +M =M 1sin 45+50=100K N ⋅m
M C ==⋅m α=63.4
β=90 γ=26. 5 6
5.4 如图所示,置于水平面上的网格,每格边长a = 1m ,力系如图所示,选O 点为简化中心,坐标如图所示。已知:F 1 = 5 N,F 2 = 4 N,F 3 = 3 N;M 1 = 4 N·m ,M 2 = 2 N·m ,求力系向O 点简化所得的主矢F 和主矩M O 。
' R
题5.4图
'
解:F R =F 1+F 2-F 3=6N
方向为Z 轴正方向
M M
x
=M 2+2F 1+2F 2-4F 3=8N ⋅m =M 1-3F 1-F 2+F 3=-12N ⋅
m
y
M O =
=14.42N ⋅m
α=56.63
β=-33.9
γ=90
5.5 解:
∑X ∑
=0, F Ax +F Bx +T 1cos 30+T 2cos 30=0
Z =0, F Az +F Bz -T 2sin 30+T 1sin 30-W =0
z
∑M ∑M ∑M
=0, -60T 1cos 30-60T 2cos 30-100F Bx =0=0, -30W +60T 1sin 30-60T 2sin 30+100F Bz =0=0, W r +T 2r 1-T 1r 1=0
x
y
F Ax =-20.78K N , F Az =13K N F Bx =7.79KN , F Bz =4.5KN T 1=10KN , T 2=5KN
5.6
题5.6图
2a ,AB 长为2b ,列出平衡方程并求解
F Bz =
F Az =
100N
5.7
y
x
题5.7图
解:∑X =0, F Ax +F Bx +F 1=0
∑Z ∑M ∑M ∑M
=0, F Az +F Bz +F =0
z
=0, -140F 1-100F Bx =0
=0,20F 1-20F =0 =0, 40F +100F Bz =0
y
x
F Ax =
320N
, F
Az =-480N
F Bx =-1120N , F Bz =-320N F =800N
5.8
题5.8图
解:G 、H 两点的位置对称于y 轴
F BG =F BH
∑X ∑Y ∑Z ∑M
=0, -F BG sin 45cos 60+F BH sin 45cos 60+F Ax =0
=0, -F BG cos 45cos 60-F BH cos 45cos 60+F Ay =0=0, F Az -F BG sin 60-F BH sin 60-W =0
x
=0, 5F BG sin 45cos 60+5F BH sin 45cos 60-5W =0
F BG =F BH =28.28K N , F Ax =0, F Ay =20K N , F Az =68.99K N 5.9
5.10。
题5.10图
解:∑Y =0, F By +F Ay =0
∑Z ∑M ∑M ∑M
=0, F Az +F Bz -F -F 1=0
x
=0, 2bF 1-cF cos α=0 =0, aF -bF Bz +bF Az =0 =0, bF By -bF Ay =0
y
z
F Ay =F By =0,F Az =423.92N , F Bz =183.92N F 1=207.84N
5.11
题 5.11图
解:三角形OAB 的中心为:(15, 6.67)
A 1=300m m 2
小圆重心为:(6, 6) A 2=16π 该薄板的重心:
x =
x 1A 1-x 2A 2
A 1-A =16.8
2
y =
y 1A 1-y 2A 2
A =-0.4
1-A 2
5.12。
x
题5.12图
解:圆重心:(0, 0)
A 2
1=14400πm m 三角形重心(0, 30)板的重心位置:
x =
x 1A 1-x 2A 2
A 1-A =0
2y
y =1A 1-y 2A 2
A 1-A =-6.54
2
x
A 2=8100m m 2
5.13
x
题5.13图
题5.14图解:I部分重心:(45, 20) ∏部分重心:(105, 20) A 22=900m m
I∏部分重心:(60, -20) A 3=4800m m 2
均质板OABCD 的重心:
x =
x 1A 1+x 2A 2+x 3A 3
A =60m m
1+A 2+A 3y 1A 1+y 2A 2+y 3A 3
y =A 2.86m m
1+A 2+A =-3
5.14
x
解:I部分重心:(45, 60), A 2
1=10800mm
∏部分重心:(73, 60), A 2=800πmm 2
A 1=2700m m 2
I∏部分重心:(45, -20), A 3=2700mm
2
均质等厚板的重心:
x =
x 1A 1+x 2A 2+x 3A 3
A 1+A 2+A 3A 1+A 2+A 3
=49.4m m
y =
y 1A 1+y 2A 2+y 3A 3
=46.5m m