both boxes slide together top box slides and bottom box...

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1) Part A) Force P is applied at a location very close to the bottom of the top box and it’s magnitude starts from zero and is slowly increased. Describe all possible cases of impending motion of the boxes if the man does not slip. The coefficient of friction at all contacting surfaces is known. (9 pts) Case 1: __both boxes slide together ________________________________________________ Case 2: __top box slides and bottom box stays in place ______________________ Case 3: __both boxes tip together about point B ________________________________ Case 4: __top box tips away from man and bottom box stays in place ___ (Case 4 will not happen unless m is very very large and weight is very very small because P is applied close to bottom of box.) Case 5 : ____________________________________________________________________________________ Case 6 : _____________________________________________________________________________________ Case 7 : ______________________________________________________________________________________ Case 8 : ______________________________________________________________________________________ Case 9 : ______________________________________________________________________________________ Case 1 0 : _____________________________________________________________________________________

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Page 1: both boxes slide together top box slides and bottom box ...web.mst.edu/~ide50-3/exams/10w/exam3/WS2010EX3.pdf · both boxes slide together ... The carpenter applies both a vertical

1) Par t A) Force P is applied at a location very close to the bottom

of the top box and it ’s magnitude star ts from zero and is slowly

increased. Descr ibe all possible cases of impending motion of

the boxes if the man does not slip. The coefficient of fr ict ion at

all contact ing sur faces is known. (9 pts)

Case 1 : __both boxes slide together________________________________________________

Case 2 : __top box slides and bottom box stays in place______________________

Case 3 : __both boxes t ip together about point B________________________________

Case 4 : __top box t ips away from man and bottom box stays in place___

(Case 4 will not happen unless m is very very large and weight is very very small because P is

applied close to bottom of box.)

Case 5 : ____________________________________________________________________________________

Case 6 : _____________________________________________________________________________________

Case 7 : ______________________________________________________________________________________

Case 8 : ______________________________________________________________________________________

Case 9 : ______________________________________________________________________________________

Case 10 : _____________________________________________________________________________________

Page 2: both boxes slide together top box slides and bottom box ...web.mst.edu/~ide50-3/exams/10w/exam3/WS2010EX3.pdf · both boxes slide together ... The carpenter applies both a vertical

Par t B) The carpenter pushes the board

hor izontally over the top of the sawhorse.

The board weighs 54 lb and the sawhorse

weighs 15 lb with a center of gravity at G.

Determine if the sawhorse will stay in

posit ion, slip, or t ip if the board is pushed

forward. Ver ify your answer by checking

ALL condit ions of impending motion and

compar ing the hor izontal force exer ted by

the carpenter , Cx. (16 pts)

Note: The carpenter applies both a ver tical and hor izontal force to keep the board in equilibr ium in

the posit ion shown.

i) if the sawhorse stays in posit ion and the board slips: Cx =__24 .3_ lb

ii) if the sawhorse slips: Cx =__19 .08 _ lb

iii) if the sawhorse t ips: Cx =__21 .2 _ lb

∴ the sawhorse will _____SLIP_____________./ 25

10 ft

C

Page 3: both boxes slide together top box slides and bottom box ...web.mst.edu/~ide50-3/exams/10w/exam3/WS2010EX3.pdf · both boxes slide together ... The carpenter applies both a vertical

2) A counterclockwise moment M = 1500 lb-in. is applied to the

flywheel. If the coefficient of fr ict ion between the band and the

wheel is 0 .20 , compute the minimum force P necessary to prevent

the wheel from rotat ing. (20 pts)

/ 20

Page 4: both boxes slide together top box slides and bottom box ...web.mst.edu/~ide50-3/exams/10w/exam3/WS2010EX3.pdf · both boxes slide together ... The carpenter applies both a vertical

3) Par t A) For the angle shape shown to the r ight,

determine the vert ical distance from point A to the

hor izontal centroidal axis. Select the best answer

from the choices below. (8 points)

(a)35 .8 mm

(b)44 .8 mm

(c)47 .2 mm Correct – Full credit

(d)53 .5 mm

(e)77 .2 mm Half credit

(f) 85 .0 mm

(g)115 .0 mm

(h)122 .8 mm Half credit

(i)125 .3 mm

(j)134 .2 mm

(k)152 .8 mm Half credit

Page 5: both boxes slide together top box slides and bottom box ...web.mst.edu/~ide50-3/exams/10w/exam3/WS2010EX3.pdf · both boxes slide together ... The carpenter applies both a vertical

Par t B) Determine for the shape to the

r ight. For maximum par t ial credit, show your

calculat ions in the tabular format below. (17

pts)

3

2

282,971.248 mm29.988 mm

9,436.283 mmy

/

Shape iA iy i iy A

Triangle 3,750 mm2 25.00 mm 93,750.000 mm

3

Middle Rectangle

6,000 mm2 37.50 mm 225,000.000 mm

3

Add-on Rectangle

1,100 mm2 47.50 mm 52,250.000 mm

3

Semicircle Cutout

–1,413.717 mm2 62.268 mm –88,028.752 mm

3

9,436.283 mm2 282,971.248 mm

3

Shape iA iy i iy A

Triangle 3,750 mm2 25.00 mm 93,750.000 mm

3

Rectangle 7,500 mm2 37.50 mm 281,250.000 mm

3

Semicircle Cutout

–1,413.717 mm2 62.268 mm –88,028.752 mm

3

Square Cutout

–400 mm2 10.00 mm –4,000.000 mm

3

9,436.283 mm2 282,971.248 mm

3

y

Page 6: both boxes slide together top box slides and bottom box ...web.mst.edu/~ide50-3/exams/10w/exam3/WS2010EX3.pdf · both boxes slide together ... The carpenter applies both a vertical

Par t B) (continued solut ion)...

Shape iA iy i iy A

Full Rec-tangle

15,000 mm2 37.50 mm 562,500.000 mm

3

Triangle Cutout

–3,750 mm2 50.00 mm –187,500.000 mm

3

Semicircle Cutout

–1,413.717 mm2 62.268 mm –88,028.752 mm

3

Square Cutout

–400 mm2 10.00 mm –4,000.000 mm

3

9,436.283 mm2 282,971.248 mm

3

Shape iA iy i iy A

Triangle 3,750 mm2 25.00 mm 93,750.000 mm

3

55-mm-tall Rectangle

5,500 mm2 47.50 mm 261,250.000 mm

3

Semicircle Cutout

–1,413.717 mm2 62.268 mm –88,028.752 mm

3

20-mm-tall Rectangle

1,600 mm2 10.00 mm 16,000.000 mm

3

9,436.283 mm2 282,971.248 mm

3

Page 7: both boxes slide together top box slides and bottom box ...web.mst.edu/~ide50-3/exams/10w/exam3/WS2010EX3.pdf · both boxes slide together ... The carpenter applies both a vertical

4) Par t A) Circle the most appropr iate answer. Pressure acts __________ to the sur face at a given

point. (4 pts)

(a)parallel

(b) normal

(c) vert ical

(d) hor izontal

Par t B) Draw the fluid pressure distr ibut ion curves on the left side of each of the two submerged

plates shown in bold below. (4 pts)

Page 8: both boxes slide together top box slides and bottom box ...web.mst.edu/~ide50-3/exams/10w/exam3/WS2010EX3.pdf · both boxes slide together ... The carpenter applies both a vertical

Par t C) The dam has its center of gravity located at

CG. Determine if the dam will be able to resist

over turning about point A. The mass density of

concrete is ρc = 2 .0 Mg/ m

3 and it ’s cross-sectional

area shown is 0 .75 m2. The width of the concrete

dam, measured into the page, is 3 m. The mass

density of water is ρw = 1 .0 M g/ m

3. The gravitat ional

constant is g = 9 .81 m/ s2. (14 pts)

/ 22