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Engineering Dynamics Homework- includes kinematics
Typology: Exercises
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Homework Assignment 2 ProblemI-
Given: ⊖ = 15 ° M- 100 kg F = 175 N To determine: Acceleration of the cart. Assume: ① Neglect friction (^2) M rope M pulleys > 0 M wheels → 0
We start by identifying some reference frames.
step 1
Step 2 : Free^ Body^ Diagram N
↑ ran
NO (^) O
a
↑ ran
NO
a
Foy
"I,T. Baecc a donnifnf heor tr op enet
the halfension
Note: He^ e^7 -^175 N which is the reaction force to the pull exerted by the man.
Step 3 : Governing equations / Equations of motion N cart Invoking N 2 L: EE = mid or, "Fg + "FT + IT 1 = 1
= M^ N^ a-^ cast
or-mg ↑ + wait III. =
M N@Cart.
IF Transformation between 'A'and 'N' babis-
i as tar ↑ 0
OR,^9 ,^ =^ coso^ it^ sino; o and ai = - sino ↑ + coso;
Revisiting eq①i
4 T (cos ⊖ ↑ + Sino;)
LHS (^) only-
↑ (-mg + Nose + 4 Tsing
OR;^ N^ a-^ car!
Another assumption- Cart does not leave contact with the incline. Therefore; A a- cart = a af + 0 %
Compute components of N a- cast in the 'A' babis- The component along ai is =
N ♂ cast. a↑
OR, C- Nsint + 4 T coso)coS⊖+(mg
ma
⇒ A^ =^4.^46 m^12
Ans
Problem- 2 - 2 Illustration^ Given: Anglee as illustrated mass of cart = M mass of cylinder = m Tofins: P such that normal reaction at B is Zero. SOLUTION Step 1 : Reference frames
m W
it
m W
it
i 'μ' °
or a
'A"
Trim W
i 'μ' °
or a
'A"
Assuming G* is the mass center of cylinder and cart.
Fg
Assume: No friction, neglect weight of wheels.
Steps: Governing equation/Equations of motion Invoking NIL: " Including forces from the FBD:-&
IF = miasys
Btfg + NT^ +^ ñ =^ may^ a-^ Sys^ (my,^ =@+^ m))
where
These are eaisni ly' (^) Ak'nbowansis.
Pai - M+m)g↑ + Nia: + Na ai = Atm) Noises Pai - (M+m)g↑ + (Nina) as = (Mtm, NEYS convert to 'N'basis Transformation across 'N'-'A' systems-
ai = cos of + sin 05 ↑ % =-sin of + coso ↑
ñ Ñ
00 f
O
Now, using transformation in g. ( 1 ) -
Now, it is more intuitive to express this acceleration in the' A' frame- if A a- Ms = a, a + as % Now, assuming the entire system stays in contact with the inclined slope always; ar
OR, A, = - IN, + Nz) Serio (M +m)
WSO + ☑ + Nz )ws^ Osino (Mtm)
OR,^ Plasoitsino;)
OR,'
[Paso^ [Psni^ o^ +^ (Nit^ Ne)^ cos^ O-M^ +^ m)^ g)^ ↑ (M+M)
N £ 87 s =
(Mtm)
OR, (^) "= (M¥m, - gsino
Steph: We also need the FBD of the cylinder → COM of the MG cylinder
i
N
r
i
j
ñ
ñ
Step 5 : Equations of motion of the cylinder Invoking our favorite NLL: IF = mage Añ 9 ᵗ OR, Fg-^ +^ NA^ tNB^ =^ mage^ Aage
OR, mgj + NA cos 30 & + Nasiri 30 a
[may = m]
Ff
A a- cyl
Express ↑ in terms of@, 4 ) → ↑ = & sin 15 + af cos 150
OR, - my sin 150 af - mg cos 15 ° at + Na cos 309 + NA bin 309 ^ 2
Now, in^ this^ particular^ case; the cylinder is part of the system. Therefore, a- eye = a- Sys (irrespective of 'A' or 'N'bases)
NB = 0 (contact lost at B)
Also,
l
2
Equating the vector equations; we have the following scalar equations - NA cos 30 °- mg sin 15 ° = M[¥m-g sin 0 ]
NA Sin 38 - mg 105150 = 0
From ⑤i NA = Mg cos/ 5 ◦ Lein 30 °
A i my cos 15 ° ✗ 805300 - mg Sin 15 = I
Sin 30 °
Substituting in / (MFM)
000 P^ =^ (Mtm)^ g^ cos^15007300 P - 1. 67303 (Mt M)g Ant