1.FBDs:shown2.Newton

G : Fx∑ = T = maGx (1)

Fy∑ = −mg = maGy (2)

A : Fx∑ = −T + F − µk N = maA (3)

Fy∑ = N − mg = 0 ⇒ N = mg (4)

3.Kinematics

!aG = !aA +

!α × !rG / A −ω 2!rG / A = aAi + α k( )× −Li( )− !0 = aAi − Lα j ⇒ i : aGx = aA (5)

4.Solve(1),(3),(4)and(5):

T = maGx = F − µkmg − maGx ⇒ aGx =

12

Fm− µk g

⎛⎝⎜

⎞⎠⎟

(2): aGy = −g Therefore,

!aG = 12

Fm− µk g⎛

⎝⎜⎞⎠⎟

i − gj

T G

F A

T

f

mg

N

mg

SOLUTION

ME 274: Basic Mechanics IISpring 2018

March 8, 2018

Problem 2 (20 points):

Given: The ladder of a fire truck rotates around the Z axis at a constant rate ⌦Z

, while rotatingupward at a constant rate ✓. The cart travels in the positive Y direction with a velocity andacceleration of V

truck

, Vtruck

, respectively. The XY Z axes are fixed. The following equation is tobe used to find the acceleration of point P on the ladder:

~a

P

= ~a

A

+�~a

P/A

�rel

+ ~↵⇥ ~r

P/A

+ 2~! ⇥�~v

P/A

�rel

+ ~! ⇥�~! ⇥ ~r

P/A

�

Find: For the positions noted in the two scenarios below, determine each of the following quantities.You may write your symbolic answers as vectors in terms of the moving xyz components OR theXY Z components, but you must be consistent between answers.

X

YZ

x

y

z

X

YZ

x

yz

1) 2)

In scenario 1), at the instant shown the movingxyz axes are aligned with the fixed XY Z axes.The observer is fixed to the rotating stage at A.

(a) ~a

A

=

(b) ~! =

(c) ~↵ =

(d)�~v

P/A

�rel

=

(e)�~a

P/A

�rel

=

In scenario 2), at the instant shown the movingxyz axes are aligned with the fixed XY Z axeswhen ✓ = 0o. The observer is fixed to ladderAP .

(a) ~a

A

=

(b) ~! =

(c) ~↵ =

(d)�~v

P/A

�rel

=

(e)�~a

P/A

�rel

=

Exam 2 Page 4 of 11

Z,z

v truc

k

Y,y

θ

ΩZ

P

A

! a A=" v tr

uckJ=" v tr

uck

j! ω=Ω

ZK=Ω

Zk

! α=" Ω

ZK+Ω

Z" K=! 0

! v P/A

() re

l=" θi ()×

dco

sθj+

dsi

nθk

()

=d" θ−s

inθ

j+co

sθk

()

=d" θ−s

inθ

J+

cosθ

K(

)! a P

/A(

) rel=−" θ

2d

cosθ

j+d

sinθ

k(

)=−" θ

2d

cosθ

J+

dsi

nθK

()

Scen

ario#1

Z

v truc

k

Y

y z

θ

ΩZ

A

P

! a A=" v tr

uckJ=" v tr

uck

cosθ

j−si

nθk

()

! ω=Ω

ZK+" θi=Ω

ZK+" θI

=Ω

Zsi

nθj+

cosθ

k(

)+" θi

! α=" Ω

ZK+Ω

Z" K+"" θi+" θ" i=" θ! ω×

i(

)=" θΩ

ZK+" θi

()×

i=" θΩ

Zsi

nθj+

cosθ

k(

)×i

=" θΩ

Zco

sθj−

sinθ

k(

)=" θΩ

ZJ

! v P/A

() re

l=! 0

! a P/A

() re

l=! 0

Scen

ario#2

ME 274: Basic Mechanics IISpring 2018

March 8, 2018

Problem 3 (16 points):

Part A (6 points) No partial credit will be given.

x

1

2

yA

P

Given: A frame of mass 2m is initially at rest on a smoothsurface. A particle P of mass m is attached to the end of arod of negligible mass. The frame is symmetric about point A.The mass pivots freely at A. If the rod is released from rest inthe horizontal position shown, i.e., at position 1. Answer thefollowing when the rod is vertical, i.e., at position 2.

Find:

1) (2 pts) When the pendulum is at position 2, circle all statements that correctly describe thespeed of the frame v

A

and speed of the pendulum v

P

:

(a) |vA

| = |vP

|,

(b) |vA

| < |vP

|,

(c) |vA

| > |vP

|,

(d) Not enough information to answer the question. Provide an explanation.

2) (2 pts) The magnitude of the force in the rod at position 2 is:

(a) F = 2mg,

(b) F = 4mg

3

,

(c) F = mg

3

,

(d) F = mg,

(e) F = 0,

(f) Not enough information to answer the question. Provide an explanation.

Exam 2 Page 7 of 11

ME 274: Basic Mechanics IISpring 2018

March 8, 2018

3) (2 pts) After it has been released the particle P continues to travel in the positive x directionfrom position 2 to the top of the frame. Circle the figure(s) that best describes the motion of thesystem as P reaches the top of the frame for the first time.

a)

b)

c)

d)

x

y

P 2

Exam 2 Page 8 of 11

ME 274: Basic Mechanics IISpring 2018

March 8, 2018

Part B (6 points) No partial credit will be given.

Given: A block of mass m is initially at rest. It is released to slide down a smooth curved ramponto the surface of a rough deck. The coe�cient of friction for the rough deck is µ

k

.

x

A

B

C

y

Smooth

Find:

1) (2 pts) In order for the block not to slide o↵ the deck, circle the statement below that must betrue:

(a) µ

k

h/L,

(b) µ

k

� h/L,

(c) Not enough information to answer the question. Provide an explanation.

Exam 2 Page 9 of 11

ME 274: Basic Mechanics IISpring 2018

March 8, 2018

x

B

C

y2) (2 pts) If the block slides on the rough surface with a giveninitial velocity v

o

, find the time it takes to stop. Assume theblock does not fall o↵ the deck. Express your answers in termsof the parameters m, v

o

, µk

, L and g. Note all parametersmay not appear in the final expression. Write your answer inthe block below.

t

stop

=

3) (2 pts) On the figure below, sketch the net impulse vector acting on the block over a period oftime t

stop

.

x

B

C

y

Exam 2 Page 10 of 11

ME 274: Basic Mechanics IISpring 2018

March 8, 2018

Part C (4 points) No partial credit will be given.

Given: A small coin of mass m is located on a horizontal surface of a disk a distance r from thecenter. The disk starts from rest and accelerates with a constant angular acceleration ↵. Thestatic coe�cient of friction between the coin and disk is µ

s

. Let ↵ = 2 rad/s, and answer thefollowing questions after �

t

= 2 seconds has elapsed.

Find:

1) (2 pts) Assume the coin has not slipped on the disk’s surface, use the figure below to sketchthe friction force acting on the coin.

2) (2 pts) Assume the coin has not slipped on the disk’s surface, use the figure below to sketchthe acceleration vector associated with the coin.

Exam 2 Page 11 of 11