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Engineering Dynamics Homework, Exercises of Engineering Dynamics

Engineering Dynamics Homework- includes kinematics

Typology: Exercises

2022/2023

Uploaded on 02/05/2025

hanlin-wang
hanlin-wang 🇺🇸

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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

O

SOLUTION :

We start by identifying some reference frames.

step 1

Step 2 : Free^ Body^ Diagram N

↑ ran

NO (^) O

a

↑ ran

NO

a

G

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

  • my ↑ + NC-sino ↑ + cos 09 ) + :(- N sino + 4 T us e) + - N Sino + UTOS- 0

4 T (cos ⊖ ↑ + Sino;)

LHS (^) only-

↑ (-mg + Nose + 4 Tsing

OR;^ N^ a-^ car!

  • mg +^ N^ Coso^ +^4 T^ sin)

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

  • Nwso + 4 Tsin a) sino = ma Nsaid OSO + 4 T sin OR, - N sino WSO^ +^4 T^ costo-mg^ sniot =^ Ma 4 T/sink + 050 ) - mg so = 012 , ma: 4 T - mg sino = 4 × 175 - 100 ×

OR,

  1. 8 ✗ sin/ 59

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

A-- 60

6 B 0 °

m W

it

P

£ 28 FREE BODY DIAGRAMS

m W

it

B

A

i 'μ' °

or a

'A"

Trim W

B

A

i 'μ' °

or a

'A"

Assuming G* is the mass center of cylinder and cart.

G

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

  • eq①

Now, using transformation in g. ( 1 ) -

  • (M+m)g 5 + N(+Nr) (-senior ↑ + cos g) = (M+m)Na→ [Paso-(Ni + Nz) said it[ 1 % 7 N'Joo so- (Mtm)g]j=(M+m)^ Natys
  • (NitNz)sino (MM)

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

  • ⑨ OR, Ay = ☑^ a-^ Sys.^ ai

OR, A, = - IN, + Nz) Serio (M +m)

WSO + ☑ + Nz )ws^ Osino (Mtm)

  • gsino + Photo + sino)

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

B

G

NB

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

  • (Np) cos 304 , + NB sin 30 at = mage

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

  • NB cos 30 ° at + NB Siri 30 ° ai = MA a- cyl O [ my sin 15 ° + Na cos 30 °- Np Los 3074 + [- mg cos 15 ° + Na bin 30 ° + N, Enzo] ai = ma a- eye

OR,

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

- B

From ⑤i NA = Mg cos/ 5 ◦ Lein 30 °

A i my cos 15 ° ✗ 805300 - mg Sin 15 = I

Sin 30 °

Substituting in / (MFM)

  • my sin 15 °

000 P^ =^ (Mtm)^ g^ cos^15007300 P - 1. 67303 (Mt M)g Ant