hw 4,me 240 dynamics, fall 2017 - 12000.org
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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
๐ฃ = 228 ๏ฟฝ1000๐๐ ๏ฟฝ ๏ฟฝ
โ3600๏ฟฝ
= (228)10003600
= 63.333 m/s
In normal direction, the acceleration is ๐๐ =๐ฃ2
๐ = (63.333)2
115 = 34.879 m/s2 or in terms of ๐, it
becomes 34.8799.81 = 3.555 g.
Now, force balance in vertical direction gives
๐ฟ โ ๐๐ = ๐๐๐๐ฟ = ๐ ๏ฟฝ๐ + ๐๐๏ฟฝ
= 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
๐๐ ๐ cos๐ + ๐ sin๐ = ๐๐ฃ2max๐
(1)
โ๐๐ + ๐ cos๐ โ ๐๐ ๐ sin๐ = 0 (2)
From (2)
๐ =๐๐
cos๐ โ ๐๐ sin๐From (1)
๐ฃ2max =๐๐
๏ฟฝ๐๐ ๐ cos๐ + ๐ sin๐๏ฟฝ
=๐๐ ๏ฟฝ๐๐ ๏ฟฝ
๐๐cos๐ โ ๐๐ sin๐๏ฟฝ
cos๐ + ๏ฟฝ๐๐
cos๐ โ ๐๐ sin๐๏ฟฝsin๐๏ฟฝ
= ๐ ๏ฟฝ๐๐ ๏ฟฝ๐
cos๐ โ ๐๐ sin๐๏ฟฝcos๐ + ๏ฟฝ
๐cos๐ โ ๐๐ sin๐๏ฟฝ
sin๐๏ฟฝ
= ๐๐ ๏ฟฝ๐๐
1 โ ๐๐ tan๐+
tan๐1 โ ๐๐ tan๐๏ฟฝ
= ๐๐ ๏ฟฝ๐๐ + tan๐1 โ ๐๐ tan๐๏ฟฝ
= 324 (9.81)
โโโโโโโ0.2 + tan ๏ฟฝ31 ๐
180๏ฟฝ
1 โ 0.2 tan ๏ฟฝ31 ๐180
๏ฟฝ
โโโโโโโ
= 2893.165 m/s
Hence
๐ฃmax = 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
โ๐๐ ๐ cos๐ + ๐ sin๐ = ๐๐ฃ2min๐
(3)
โ๐๐ + ๐ cos๐ + ๐๐ ๐ sin๐ = 0 (4)
From (4)
๐ =๐๐
cos๐ + ๐๐ sin๐
3
From (3)
๐ฃ2min =๐๐
๏ฟฝโ๐๐ ๐ cos๐ + ๐ sin๐๏ฟฝ
=๐๐ ๏ฟฝโ๐๐ ๏ฟฝ
๐๐cos๐ + ๐๐ sin๐๏ฟฝ
cos๐ + ๏ฟฝ๐๐
cos๐ + ๐๐ sin๐๏ฟฝsin๐๏ฟฝ
= ๐๐ ๏ฟฝโ๐๐ ๏ฟฝ1
cos๐ + ๐๐ sin๐๏ฟฝcos๐ + ๏ฟฝ
1cos๐ + ๐๐ sin๐๏ฟฝ
sin๐๏ฟฝ
= ๐๐ ๏ฟฝโ๐๐
1 + ๐๐ tan๐+
tan๐1 + ๐๐ tan๐๏ฟฝ
= ๐๐ ๏ฟฝtan๐ โ ๐๐ 1 + ๐๐ tan๐๏ฟฝ
= 324 (9.81)
โโโโโโโ
tan ๏ฟฝ31 ๐180
๏ฟฝ โ 0.2
1 + 0.2 tan ๏ฟฝ31 ๐180
๏ฟฝ
โโโโโโโ
= 1137.425 m/s
Hence
๐ฃmin = 33.726 m/s
= 33.726 (0.001) (3600)= 121.414 km/hr
Hence
121.414 โค ๐ฃ โค 193.637
0.3 Problem 3
4
0.3.1 part (1)
Since acceleration is constant along barrel, then using
๐ฃ2๐ = ๐ฃ20 + 2๏ฟฝฬ๏ฟฝ๐ฟ
We will solve for ๏ฟฝฬ๏ฟฝ. Assuming ๐ฃ0 = 0 then
16812 = 2๏ฟฝฬ๏ฟฝ (4.2)
๏ฟฝฬ๏ฟฝ =16812
2 (4.2)= 336400.1 m/s2
But
๏ฟฝฬ๏ฟฝ = ๏ฟฝ๏ฟฝฬ๏ฟฝ โ ๐ฟ๏ฟฝฬ๏ฟฝ2๏ฟฝ ๏ฟฝฬ๏ฟฝ๐ + ๏ฟฝ๐ฟ๏ฟฝฬ๏ฟฝ + 2๏ฟฝฬ๏ฟฝ๏ฟฝฬ๏ฟฝ๏ฟฝ ๏ฟฝฬ๏ฟฝ๐But ๏ฟฝฬ๏ฟฝ = 0, hence the above becomes
๏ฟฝฬ๏ฟฝ = ๏ฟฝ336400.1 โ (4.2) (0.18)2๏ฟฝ ๏ฟฝฬ๏ฟฝ๐ + (2 (1681) (0.18)) ๏ฟฝฬ๏ฟฝ๐= 336400 ๏ฟฝฬ๏ฟฝ๐ + 605.16 ๏ฟฝฬ๏ฟฝ๐
0.3.2 part (2)
Free body diagram for bullet gives
๐ = ๐๐๐ + ๐๐ sin๐
but ๐๐ = 336400 m/s2 and ๐ = 16.9 kg and ๐ = 230, hence
๐ = (16.9) (336400) + (16.9) (9.81) sin ๏ฟฝ(23)๐180
๏ฟฝ
= 5685225 N= 5.685 ร 106 N
0.3.3 Part (3)
Free body diagram for bullet gives
๐ = ๐๐๐ + ๐๐ cos๐
= (16.9) (605.16) + (16.9) (9.81) cos ๏ฟฝ(23) ๐180
๏ฟฝ
= 10379.81 N= 10.379 ร 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
โโ
A
B
maAx
maAy = 0
maBx
maBy = 0
For ๐ด,๏ฟฝ๐น๐ฅ = ๐๐ด๐๐ด๐ฅ
๐น๐ด = ๐๐ด๐๐ด๐ฅ (2)
๏ฟฝ๐น๐ฆ = ๐๐ด๐๐ด๐ฆ
๐๐ด โ๐๐ด = 0 (3)
For ๐ต๏ฟฝ๐น๐ฅ = ๐๐ต๐๐ต๐ฅ
๐น๐ โ ๐น๐ด โ ๐น๐ต = ๐๐ต๐๐ต๐ฅ (4)
๏ฟฝ๐น๐ฆ = ๐๐ต๐๐ต๐ฆ๐๐ต โ๐๐ต โ ๐๐ด = 0 (5)
Hence (3) becomes
๐๐ด = ๐๐ด (6A)
= ๐๐ด๐
And (5) becomes
๐๐ต = ๐๐ต + ๐๐ด
= ๐๐ต๐ + ๐๐ด๐= (๐๐ต + ๐๐ด) ๐ (6B)
But ๐น๐ด = ๐๐ด๐1 then (2) becomes
๐๐ด๐1 = ๐๐ด๐๐ด๐ฅ
But from (6) the above reduces to (since ๐๐ด = ๐๐ด๐)
๐๐ด๐๐1 = ๐๐ด๐๐ด๐ฅ
6
Hence
๐๐ด๐ฅ = ๐๐1
= (32.2) (0.26)
= 8.372 ft/s2
Similarly ๐น๐ต = ๐๐ต๐2 hence (4) becomes
๐น๐ โ ๐๐ด๐1 โ ๐๐ต๐2 = ๐๐ต๐๐ต๐ฅBut ๐๐ต = (๐๐ด + ๐๐ต) ๐, then above becomes
๐น๐ โ ๐๐ด๐๐1 โ (๐๐ด + ๐๐ต) ๐๐2 = ๐๐ต๐๐ต๐ฅ
๐๐ต๐ฅ =๐น๐ โ ๐๐ด๐๐1 โ (๐๐ด + ๐๐ต) ๐๐2
๐๐ต
๐๐ต๐ฅ =438 โ (50) (0.26) โ (50 + 71) (0.46)
7132.2
= 167.504 ft/sec2
0.5 Problem 5
Free body diagram for block ๐ต results in
2๐ โ8๐๐๐ต
= ๐๐ด
Free body diagram for block ๐ด resuts in
โ๐๐ด๐ sin๐ โ ๐๐๐ด๐ cos๐ + 2๐ = ๐๐ด๐๐ดIn addision, since rope length is fixed, we find that ๐๐ด = 2๐๐ต. The above are 3 equations in3 unknowns ๐, ๐๐ด, ๐๐ต. Solving gives
๐๐ด = 4.439 m/s2
๐๐ต = 2.2197 m/s2
๐ = 37.95 N
7
0.6 Problem 6
maximum displacement is 2 ft. Maximum speed is 9.404 ft/sec.
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