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HW 4,ME 240 Dynamics, Fall 2017
Nasser M. Abbasi
December 30, 2019
0.1 Problem 1
The velocity in meter per second is
π£ = 228 οΏ½1000ππ οΏ½ οΏ½
β3600οΏ½
= (228)10003600
= 63.333 m/s
In normal direction, the acceleration is ππ =π£2
π = (63.333)2
115 = 34.879 m/s2 or in terms of π, it
becomes 34.8799.81 = 3.555 g.
Now, force balance in vertical direction gives
πΏ β ππ = ππππΏ = π οΏ½π + πποΏ½
= 970 (9.81 + 34.879)= 43348.33 N
0.2 Problem 2
1
2
For maximum speed, friction acts downwards as car assumed to be just about to slideupwards. Resolving forces in normal and tangential gives
ππ π cosπ + π sinπ = ππ£2maxπ
(1)
βππ + π cosπ β ππ π sinπ = 0 (2)
From (2)
π =ππ
cosπ β ππ sinπFrom (1)
π£2max =ππ
οΏ½ππ π cosπ + π sinποΏ½
=ππ οΏ½ππ οΏ½
ππcosπ β ππ sinποΏ½
cosπ + οΏ½ππ
cosπ β ππ sinποΏ½sinποΏ½
= π οΏ½ππ οΏ½π
cosπ β ππ sinποΏ½cosπ + οΏ½
πcosπ β ππ sinποΏ½
sinποΏ½
= ππ οΏ½ππ
1 β ππ tanπ+
tanπ1 β ππ tanποΏ½
= ππ οΏ½ππ + tanπ1 β ππ tanποΏ½
= 324 (9.81)
βββββββ0.2 + tan οΏ½31 π
180οΏ½
1 β 0.2 tan οΏ½31 π180
οΏ½
βββββββ
= 2893.165 m/s
Hence
π£max = 53.788 m/s
= 53.788 (0.001) (3600)= 193.637 km/hr
For minimum speed, friction acts upwards as car assumed to be just about to slide down-wards. Resolving forces in normal and tangential gives
βππ π cosπ + π sinπ = ππ£2minπ
(3)
βππ + π cosπ + ππ π sinπ = 0 (4)
From (4)
π =ππ
cosπ + ππ sinπ
3
From (3)
π£2min =ππ
οΏ½βππ π cosπ + π sinποΏ½
=ππ οΏ½βππ οΏ½
ππcosπ + ππ sinποΏ½
cosπ + οΏ½ππ
cosπ + ππ sinποΏ½sinποΏ½
= ππ οΏ½βππ οΏ½1
cosπ + ππ sinποΏ½cosπ + οΏ½
1cosπ + ππ sinποΏ½
sinποΏ½
= ππ οΏ½βππ
1 + ππ tanπ+
tanπ1 + ππ tanποΏ½
= ππ οΏ½tanπ β ππ 1 + ππ tanποΏ½
= 324 (9.81)
βββββββ
tan οΏ½31 π180
οΏ½ β 0.2
1 + 0.2 tan οΏ½31 π180
οΏ½
βββββββ
= 1137.425 m/s
Hence
π£min = 33.726 m/s
= 33.726 (0.001) (3600)= 121.414 km/hr
Hence
121.414 β€ π£ β€ 193.637
0.3 Problem 3
4
0.3.1 part (1)
Since acceleration is constant along barrel, then using
π£2π = π£20 + 2οΏ½ΜοΏ½πΏ
We will solve for οΏ½ΜοΏ½. Assuming π£0 = 0 then
16812 = 2οΏ½ΜοΏ½ (4.2)
οΏ½ΜοΏ½ =16812
2 (4.2)= 336400.1 m/s2
But
οΏ½ΜοΏ½ = οΏ½οΏ½ΜοΏ½ β πΏοΏ½ΜοΏ½2οΏ½ οΏ½ΜοΏ½π + οΏ½πΏοΏ½ΜοΏ½ + 2οΏ½ΜοΏ½οΏ½ΜοΏ½οΏ½ οΏ½ΜοΏ½πBut οΏ½ΜοΏ½ = 0, hence the above becomes
οΏ½ΜοΏ½ = οΏ½336400.1 β (4.2) (0.18)2οΏ½ οΏ½ΜοΏ½π + (2 (1681) (0.18)) οΏ½ΜοΏ½π= 336400 οΏ½ΜοΏ½π + 605.16 οΏ½ΜοΏ½π
0.3.2 part (2)
Free body diagram for bullet gives
π = πππ + ππ sinπ
but ππ = 336400 m/s2 and π = 16.9 kg and π = 230, hence
π = (16.9) (336400) + (16.9) (9.81) sin οΏ½(23)π180
οΏ½
= 5685225 N= 5.685 Γ 106 N
0.3.3 Part (3)
Free body diagram for bullet gives
π = πππ + ππ cosπ
= (16.9) (605.16) + (16.9) (9.81) cos οΏ½(23) π180
οΏ½
= 10379.81 N= 10.379 Γ 103 N
0.4 Problem 4
5
The free body diagram is
A
BF0
A
BF0
FBD
WA
FA
NA
WB NA
FA
FB
NB
ββ
A
B
maAx
maAy = 0
maBx
maBy = 0
For π΄,οΏ½πΉπ₯ = ππ΄ππ΄π₯
πΉπ΄ = ππ΄ππ΄π₯ (2)
οΏ½πΉπ¦ = ππ΄ππ΄π¦
ππ΄ βππ΄ = 0 (3)
For π΅οΏ½πΉπ₯ = ππ΅ππ΅π₯
πΉπ β πΉπ΄ β πΉπ΅ = ππ΅ππ΅π₯ (4)
οΏ½πΉπ¦ = ππ΅ππ΅π¦ππ΅ βππ΅ β ππ΄ = 0 (5)
Hence (3) becomes
ππ΄ = ππ΄ (6A)
= ππ΄π
And (5) becomes
ππ΅ = ππ΅ + ππ΄
= ππ΅π + ππ΄π= (ππ΅ + ππ΄) π (6B)
But πΉπ΄ = ππ΄π1 then (2) becomes
ππ΄π1 = ππ΄ππ΄π₯
But from (6) the above reduces to (since ππ΄ = ππ΄π)
ππ΄ππ1 = ππ΄ππ΄π₯
6
Hence
ππ΄π₯ = ππ1
= (32.2) (0.26)
= 8.372 ft/s2
Similarly πΉπ΅ = ππ΅π2 hence (4) becomes
πΉπ β ππ΄π1 β ππ΅π2 = ππ΅ππ΅π₯But ππ΅ = (ππ΄ + ππ΅) π, then above becomes
πΉπ β ππ΄ππ1 β (ππ΄ + ππ΅) ππ2 = ππ΅ππ΅π₯
ππ΅π₯ =πΉπ β ππ΄ππ1 β (ππ΄ + ππ΅) ππ2
ππ΅
ππ΅π₯ =438 β (50) (0.26) β (50 + 71) (0.46)
7132.2
= 167.504 ft/sec2
0.5 Problem 5
Free body diagram for block π΅ results in
2π β8πππ΅
= ππ΄
Free body diagram for block π΄ resuts in
βππ΄π sinπ β πππ΄π cosπ + 2π = ππ΄ππ΄In addision, since rope length is fixed, we find that ππ΄ = 2ππ΅. The above are 3 equations in3 unknowns π, ππ΄, ππ΅. Solving gives
ππ΄ = 4.439 m/s2
ππ΅ = 2.2197 m/s2
π = 37.95 N
7
0.6 Problem 6
maximum displacement is 2 ft. Maximum speed is 9.404 ft/sec.