07 Exercise Solutions e - ywgs.edu.hk (More About Force) … ·...
5
0.9 T 1 (40 × 9.81)(1) = (T sin 45 )(2) ∴ T ≈ 277 277 θ = 180 − 45 2 = 67.5 (40 × 9.81 × sin 45 )(1) = (T sin 67.5 )(2) ∴ T ≈ 150 150 P Fd YP = 600d XP F d YP d XP = 600 F (15) = 600 ∴ F = 40 40 Q P Q P Q P Q P A B B C 1 −y 1 +y F T = W cos 30 < W F F = W sin 30 cos 30 < sin 30 F < T
Transcript of 07 Exercise Solutions e - ywgs.edu.hk (More About Force) … ·...
0.9
T
1
(40×9.81)(1) = (T sin45 )(2)
∴ T ≈ 277
277
θ =
180 −45
2= 67.5
(40×9.81× sin45 )(1) = (T sin67.5 )(2)
∴ T ≈ 150
150
P
F dY P = 600dX P
F
(
dY P
dX P
)
= 600
F (15) = 600
∴ F = 40
40
Q P
Q
P Q
P Q
P
A B
B
C
1
−y 1
+y
F
T =W cos30 <W
F
F =W sin30
cos30 < sin30 F < T
X
X
X
Y
R +F sinθ =W ⇒ R <W
R W
ℓ
��mg (3ℓ) =��mg (10−ℓ)+��mg (5−ℓ) ⇒ ℓ= 3
P
M g ×ℓ= F sinθ×1.5
∴ F =
M gℓ
1.5sinθ◦
ℓ= 1 θ = 45
F =
M g (1)
1.5sin45◦≈ 0.94M g
ℓ θ
F
F < 0.94M g
F = ma
F1 −F2 = (3m)a ⇒ a =
F1 −F2
3m
P
F = ma
F1 −T = ma
T
T = F1 −��m
(
F1 −F2
3��m
)
=
2F1 +F2
3
=
2F1 +F2
3
W = 30sinθ+20sinφ
< 30+20sinφ (∵ sinθ < 1 0 < θ < 90 )
< 30+20 (∵ sinφ< 1 0 <φ< 90 )
W < 50
M =
(
1.2×105)
+2×(
6×104)
= 2.4×105
F = ma
F − f1 −2 f2 = M a
106−
(
3.6×105)
−2(
1.8×105)
=
(
2.4×105)
a
∴ a = 1.167
≈ 1.17−2
1.17 −2
T1 T2
F = ma
F −T1 − f1 = m1a
106−T1 −
(
3.6×105)
=
(
1.2×105)
(1.167)
T1 = 5×105
F = ma
T2 − f2 = m2a
T2 −
(
1.8×105)
=
(
6×104)
(1.167)
T2 = 2.5×105
5×105 2.5×105
F −T1 − f1 = ma → T1 = F − f1 −ma
a
F
a
F
F a
m
F = ma
mB g +mC g − f = (mA +mB +mC )a
(1)(9.81)+ (1)(9.81)−4 = (1+1+1)a
∴ a = 5.207 ≈ 5.21−2
5.21 −2
A F = ma
TAB − f = mA a
TAB −4 = (1)(5.207)
∴ TAB ≈ 9.21
C F = ma
mC g −TBC = mC a
(1)(9.81)−TBC = (1)(5.207)
∴ TBC ≈ 4.60
A B 9.21
B C 4.60
T cos10 = mg
T cos10 = (2000)(9.81)
∴ T = 1.992×104≈ 1.99×10
4
1.99×104
T sin10 = ma(
1.992×104)
sin10 = 2000a
∴ a = 1.730 ≈ 1.73−2
1.73 −2
L sinθ−T sin10 = M a
L sinθ = M a +T sin10
= (2500)(1.730)+(
1.992×104)
sin10
= 7784
L cosθ = M g +T cos10
= (2500)(9.81)+19923cos10
= 4.415×104
L
L =
√
77842+
(
4.415×104)2
≈ 4.483×104
L
tanθ =
7784
4.415×104⇒ θ ≈ 10.0
44800
10.0
F = mg
θ = 30
T = 2(100cos30 ) ≈ 173
θ = ϕ T
F
mg
A
RA = mg cosθ
= (0.5)(9.81)cos15
= 4.737 ≈ 4.74
A
f A = mg sinθ
= (0.5)(9.81)sin15
= 1.270 ≈ 1.27
A B 4.74 1.27
F = ma
f = M g sin45
= (0.5+1)(9.81)sin45
≈ 10.4
B
10.4
A F = ma
f = mg sin60
= (0.5)(9.81)sin60
≈ 4.25
4.25
F W
F =W sinθ
θ = 90 F = 700
W =
700
sin90= 700
θ = 20
θ = 20 f > F
f ′= 240
0
R
W cosθ
R =W cosθ
= 700cos20
= 657.8 ≈ 658
R 658
θ > 30 F
f
3
FA FB
A B
B
mg
(
L
2
)
= FAdAB
(80×9.81)×5
2= FA ×4.2
∴ FA = 467.1 ≈ 467
FA +FB = mg
467.1+FB = (80)(9.81)
∴ FB ≈ 318
A B
467 318
![5HFRQVWUXFWLRQ 3URWRFRO - Orthopedic ONE · 2015. 1. 21. · $&/ 5hfrqvwuxfwlrq 3urwrfro zlwk phqlvfdo uhsdlu &217,18(' 3odq &rqwlqxh skdvh , h[huflvhv &rqwlqxh sdwhoodu prelol]dwlrq](https://static.fdocuments.net/doc/165x107/5ff1fc0869cdfa5a3e075acf/5hfrqvwuxfwlrq-3urwrfro-orthopedic-one-2015-1-21-5hfrqvwuxfwlrq-3urwrfro.jpg)
![&KDSWHU *UDSKV · 2018-03-15 · } Ç ]pz îìíòw }v µ ]}vu/v x &kdswhu *udskv 6hfwlrq ŠŠ = =() 6lqfhwkhvxpriwkhvtxduhvr iwzrriwkhvlghv riwkhwuldqjohhtxdo vwkhvtxduhriwkhwklugvlgh](https://static.fdocuments.net/doc/165x107/5ec050b1a31ad72fa04e7c22/kdswhu-udskv-2018-03-15-pz-w-v-vuv-x-kdswhu-udskv.jpg)
