Mechanical Engineering Chapter 8 Homework The Coefficient Static Friction Between His Shoes

*8–40.

If determine the minimum coefficient of static

friction at Aand Bso that equilibrium of the supporting

frame is maintained regardless of the mass of the cylinder C.

Neglect the mass of the rods.

u

=30° C

LL

uu

SOLUTION

Ans:

789

8–41.

If the coefficient of static friction at Aand Bis

determine the maximum angle so that the frame remains

in equilibrium, regardless of the mass of the cylinder.

Neglect the mass of the rods.

SOLUTION

Free-Body Diagram:Due to the symmetrical loading and system, ends Aand Bof

Equations of Equilibrium:We have

u

ms

=0.6, C

LL

uu

Ans:

790

8–42.

The 100-kg disk rests on a surface for which m

B=0.2.

Determine the smallest vertical force P that can be applied

tangentially to the disk which will cause motion to impend.

Ans:

SOLUTION

Equations of Equilibrium. Referring to the FBD of the disk shown in Fig. a,

0.5 m

B

A

P

791

8–43.

Investigate whether the equilibrium can be maintained. The

uniform block has a mass of 500 kg, and the coefficient of

static friction is m

s=0.3.

SOLUTION

Equations of Equilibrium. The block would move only if it slips at corner O.

Referring to the FBD of the block shown in Fig. a,

A

800 mm

200 mm

3

45

600 mm

B

*8–44.

The homogenous semicylinder has a mass of 20 kg and mass

center at G. If force P is applied at the edge, and r

=

300mm,

determine the angle

u

at which the semicylinder is on the

verge of slipping. The coefficient of static friction between

the plane and the cylinder is m

s=0.3.

Also, what is the

corresponding force P for this case?

G

u

P

r

4r

3p

SOLUTION

Equations of Equilibrium. Referring to the FBD of the semicylinder shown in Fig. a,

Friction. Since the semicylinder is required to be on the verge to slip at point A,

Solving by trial and error

793

8–45.

SOLUTION

Member AB:

Post:

T

h

e

b

eam AB

h

as a neg

li

g

ibl

e mass an

d

t

hi

c

k

ness an

d

i

s

subjected to a triangular distributed loading. It is supported

at one end by a pin and at the other end by a post having a

mass of 50 kg and negligible thickness. Determine the

minimum force Pneeded to move the post.The coefficients

of static friction at Band Care and

respectively.

mC=0.2,mB=0.4

2m 400 mm

800 N/m

C

B

300 mm

A

P

4

3

5

Ans:

794

8–46.

T

h

e

b

eam AB

h

as a neg

li

g

ibl

e mass an

d

t

hi

c

k

ness an

d

i

s

subjected to a triangular distributed loading. It is supported

at one end by a pin and at the other end by a post having a

mass of 50 kg and negligible thickness. Determine the two

coefficients of static friction at Band at Cso that when the

magnitude of the applied force is increased to

the post slips at both Band Csimultaneously.

P=150 N,

SOLUTION

Member AB:

Post:

2m 400 mm

800 N/m

C

B

300 mm

A

P

4

3

5

Ans:

795

8–47.

B

AC

D

SOLUTION

Equations of Equilibrium. Referring to Fig. a,

Crates Aand Bweigh 200 lb and 150 lb, respectively.They

are connected together with a cable and placed on the

inclined plane. If the angle is gradually increased,

determine when the crates begin to slide.The coefficients

of static friction between the crates and the plane are

and .mB=0.35mA=0.25

u

796

*8–48.

Two blocks A and B, each having a mass of 5 kg, are

connected by the linkage shown. If the coefficient of static

frictionat the contacting surfaces is m

s=0.5

, determine the

largest force P that can be applied to pin C of the linkage

without causing the blocks to move. Neglect the weight of

the links.

SOLUTION

Equations of Equilibrium. Analyze the equilibrium of Joint C Fig. a,

P

30

A

C

B

797

8–49.

The uniform crate has a mass of 150 kg. If the coefficient of

static friction between the crate and the floor is m

s=0.2

,

determine whether the 85-kg man can move the crate. The

coefficient of static friction between his shoes and the floor

is m

′

s=0.4

. Assume the man only exerts a horizontal force

on the crate.

SOLUTION

Equations of Equilibrium. Referring to the FBD of the crate shown in Fig. a,

2.4 m

1.2 m

1.6 m

798

8–50.

The uniform crate has a mass of 150 kg. If the coefficient of

static friction between the crate and the floor is m

s=0.2

,

determine the smallest mass of the man so he can move the

crate. The coefficient of static friction between his shoes and

the floor is m

′

s=0.45

. Assume the man exerts only a

horizontal force on the crate.

SOLUTION

Equations of Equilibrium. Referring to the FBD of the crate shown in Fig. a,

2.4 m

1.2 m

1.6 m

799

8–51.

Beam AB has a negligible mass and thickness, and supports

the 200-kg uniform block. It is pinned at A and rests on the

top of a post, having a mass of 20 kg and negligible thickness.

Determine the minimum force P needed to move the post.

The coefficients of static friction at B and C are m

B=0.4

and m

C=0.2

, respectively.

SOLUTION

Equations of Equilibrium. Referring to the FBD of member AB shown in Fig. a,

Then consider the FBD of member BC shown in Fig. b,

1.5 m 1.5 m

C

B

0.75 m

1 m

AP

4

3

5

800

*8–52.

Beam AB has a negligible mass and thickness, and supports

the 200-kg uniform block. It is pinned at A and rests on the

top of a post, having a mass of 20 kg and negligible thickness.

Determine the two coefficients of static friction at B and atC

so that when the magnitude of the applied force is increased

to

P=300 N,

the post slips at both B and C simultaneously.

SOLUTION

Equations of Equilibrium. Referring to the FBD of member AB shown in Fig. a,

Friction. It is required that slipping occurs at B and simultaneously. Then

1.5 m 1.5 m

C

B

0.75 m

1 m

AP

4

3

5

801

8–53.

Determine the smallest couple moment that can be applied

to the 150-lb wheel that will cause impending motion. The

uniform concrete block has a weight of 300 lb. The

coefficients of static friction are m

A=0.2

, m

B=0.3

, and

between the concrete block and the floor, m

=0.4

.

1 ft

5 ft

B

A

1.5 ft

M

Ans:

SOLUTION

Equations of Equilibrium. Referring to the FBD of the concrete block, Fig. a.

Also, from the FBD of the wheel, Fig. b.

Friction. Assuming that the impending motion is caused by the rotation of wheel

due to the slipping at A and B. Thus,

Solving Eqs. (1) to (8),

8–54.

AB

C

G

2.5 m

0.25 m

2.5 m

u

Determine the greatest angle so that the ladder does not

slip when it supports the 75-kg man in the position shown.

The surface is rather slippery, where the coefficient of static

friction at Aand Bis .

SOLUTION

Free-Body Diagram:The slipping could occur at either end Aor Bof the ladder.We

Equations of Equilibrium:Referring to the free-body diagram shown in Fig.b,

we have

Dividing Eq. (1) by Eq. (2) yields

ms=0.3

X

Ans:

8–55.

The wheel weighs 20 lb and rests on a surface for which

mB=0.2. A cord wrapped around it is attached to the top

of the 30-lb homogeneous block. If the coefﬁ cient of static

friction at D is mD=0.3, determine the smallest vertical

force that can be applied tangentially to the wheel which

will cause motion to impend.

SOLUTION

Cylinder A:

1.5 ft

C

DB

A

P

3 ft

804

*8–56.

The disk has a weight Wand lies on a plane which has a

coefficient of static friction . Determine the maximum

height hto which the plane can be lifted without causing the

disk to slip.

m

SOLUTION

Unit Vector: The unit vector perpendicular to the inclined plane can be determined

using cross product.

Thus

Divide Eq. (2) by (1) yields

z

x

y

2a

a

h

805

8–57.

SOLUTION

The man has a weight of 200 lb, and the coefficient of static

friction between his shoes and the floor is

Determine where he should position his center of gravity G

at din order to exert the maximum horizontal force on the

door.What is this force?

ms=0.5.

G

Ans:

806

8–58.

Determine the largest angle that will cause the wedge to

be self-locking regardless of the magnitude of horizontal

force Papplied to the blocks.The coefficient of static

friction between the wedge and the blocks is .

Neglect the weight of the wedge.

SOLUTION

ms=0.3

X

PP

u

8–59.

SOLUTION

Equations of Equilibrium and Friction: If the wedge is on the verge of moving to

From FBD (b),

If the beam AD is loaded as shown, determine the

horizontal force Pwhich must be applied to the wedge in

order to remove it from under the beam.The coefficients of

static friction at the wedge’s top and bottom surfaces are

and respectively. If is the

wedge self-locking? Neglect the weight and size of the

wedge and the thickness of the beam.

P=0,mCB =0.35,mCA =0.25

3m

AP

10°

4 kN/m

C

B

4m

D

Ans: