μs Nc - TaiLieu.VN

Engineering Mechanics - Statics Chapter 8

Check: If FA 604 N= < FAmax 664 N=

then our no-slip assumption is good.

Problem 8-10

The block brake is used to stop the wheel from rotating when the wheel is subjected to acouple moment M0 If the coefficient of static friction between the wheel and the block is μs,determine the smallest force P that should be applied.

Solution:

ΣMC = 0; P a N b− μs Nc+ 0=

NP a

b μs c−=

μs Nr MO− 0=ΣMO = 0;

μs P a r

b μs c−MO=

PMO b μs c−( )

μs r a=

Problem 8-11

The block brake is used to stop the wheel from rotating when the wheel is subjected to a couple

771

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material may

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moment M0 If the coefficient of static friction between the wheel and the block is μs , show that the

brake is self locking, i. e., P 0≤ , provided bc

μs≤

Solution:

ΣMC = 0; P a N b− μs Nc+ 0=

NP a

b μs c−=

ΣMO = 0; μs Nr MO− 0=

μs P a r

b μs c−MO=

PMO b μs c−( )

μs r a=

P < 0 if b μs c−( ) 0< i.e. if bc

μs<

Problem 8-12

The block brake is used to stop the wheel from rotating when the wheel is subjected to a couplemoment M0 If the coefficient of static friction between the wheel and the block is μs , determine thesmallest force P that should be applied if the couple moment MO is applied counterclockwise.

772




Solution:

ΣMC = 0; P a N b− μs N c− 0=

NP a

b μs c+=

ΣMO = 0; μs− N r MO+ 0=

μs P a r

b μs c+MO=

PMO b μs c+( )

μs ra=

Problem 8-13

The block brake consists of a pin-connected lever and friction block at B. The coefficient of staticfriction between the wheel and the lever is μs and a torque M is applied to the wheel. Determine ifthe brake can hold the wheel stationary when the force applied to the lever is (a) P1 (b) P2.

773




μs 0.3=

M 5 N m⋅=

a 50 mm=

b 200 mm=

c 400 mm=

r 150 mm=

P1 30 N=

P2 70 N=

Solution: To hold lever:

ΣMO = 0; FB r M− 0=

FBMr

= FB 33.333 N=

Require NBFB

μs= NB 111.1 N=

Lever,

ΣMA = 0; PReqd b c+( ) NB b− FB a− 0=

PReqdNB b FB a+

b c+= PReqd 39.8 N=

(a) If P1 30.00 N= > PReqd 39.81 N= then the break will hold the wheel

(b) If P2 70.00 N= > PReqd 39.81 N= then the break will hold the wheel

Problem 8-14

The block brake consists of a pin-connected lever and friction block at B. The coefficient of staticfriction between the wheel and the lever is μs and a torque M is applied to the wheel. Determine ifthe brake can hold the wheel stationary when the force applied to the lever is (a) P1 (b) P2.

Assume that the torque M is applied counter-clockwise.

774

Given:




μs 0.3=

M 5 N m⋅=

a 50 mm=

b 200 mm=

c 400 mm=

r 150 mm=

P1 30 N=

P2 70 N=

Solution: To hold lever:

ΣMO = 0; FB r M− 0=

FBMr

= FB 33.333 N=

Require NBFB

μs= NB 111.1 N=

Lever,

ΣMA = 0; PReqd b c+( ) NB b− FB a+ 0=

PReqdNB b FB a−

b c+= PReqd 34.3 N=

(a) If P1 30.00 N= > PReqd 34.26 N= then the break will hold the wheel

(b) If P2 70.00 N= > PReqd 34.26 N= then the break will hold the wheel

775

Given:




Problem 8-15

The doorstop of negligible weight is pin connectedat A and the coefficient of static friction at B is μs.Determine the required distance s from A to thefloor so that the stop will resist opening of thedoor for any force P applied to the handle.

Given:

μs 0.3=

a 1.5 in=

Solution:

ΣFy = 0; NBs

s2 a2+

⎛⎜⎝

⎞⎟⎠

FA− 0=

ΣFx = 0; μs NBa

s2 a2+

⎛⎜⎝

⎞⎟⎠

FA− 0=

μs s

s2 a2+

⎛⎜⎜⎝

⎞⎟⎟⎠

FAa

s2 a2+

⎛⎜⎝

⎞⎟⎠

FA− 0=

μs s a= saμs

= s 5.00 in=

Problem 8-16

The chair has a weight W and center of gravity at G. It is propped against the door as shown.If the coefficient of static friction at A is μA, determine the smallest force P that must beapplied to the handle to open the door.

776




Given:

μA 0.3=

a 1.20 ft=

b 0.75 ft=

c 3ft=

θ 30 deg=

W 10 lb=

Solution:

Guesses By 1 lb= NA 1 lb= P 1 lb=

Given

ΣFx = 0; P− μA NA+ 0=

ΣFy = 0; NA W− By− 0=

ΣΜΒ = 0; μA NA c cos θ( ) NA c sin θ( )− W c a−( )sin θ( ) b cos θ( )+⎡⎣ ⎤⎦+ 0=

By

NA

P

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

Find By NA, P,( )=

By 11.5 lb=

NA 21.5 lb=

P 6.45 lb=

Problem 8-17

The uniform hoop of weight W is suspended from the peg at A and a horizontal force P isslowly applied at B. If the hoop begins to slip at A when the angle is θ , determine thecoefficient of static friction between the hoop and the peg.

777




Given:

θ 30 deg=

Solution:

ΣFx = 0; μ NA cos θ( ) P+ NA sin θ( )− 0=

P μ cos θ( ) sin θ( )−( )NA=

ΣFy = 0; μ NA sin θ( ) W− NA cos θ( )+ 0=

W μ sin θ( ) cos θ( )+( )NA=

ΣΜΑ = 0; W− r sin θ( ) P r r cos θ( )+( )+ 0=

W sin θ( ) P 1 cos θ( )+( )=

μ sin θ( ) cos θ( )+( )sin θ( ) sin θ( ) μ cos θ( )−( ) 1 cos θ( )+( )=

μsin θ( )

1 cos θ( )+= μ 0.27=

Problem 8-18

The uniform hoop of weight W is suspended from the peg at A and a horizontal force P isslowly applied at B. If the coefficient of static friction between the hoop and peg is μs,determine if it is possible for the hoop to reach an angle θ before the hoop begins to slip.

778




Given:

μs 0.2=

θ 30 deg=

Solution:

ΣFx = 0; μ NA cos θ( ) P+ NA sin θ( )− 0=

P μ cos θ( ) sin θ( )−( ) NA=

ΣFy = 0; μ NA sin θ( ) W− NA cos θ( )+ 0=

W μ sin θ( ) cos θ( )+( ) NA=

ΣΜΑ = 0; W− r sin θ( ) P r r cos θ( )+( )+ 0=

W sin θ( ) P 1 cos θ( )+( )=

μ sin θ( ) cos θ( )+( )sin θ( ) sin θ( ) μ cos θ( )−( ) 1 cos θ( )+( )=

μsin θ( )

1 cos θ( )+= μ 0.27=

If μs 0.20= < μ 0.27= then it is not possible to reach θ 30.00 deg= .

Problem 8-19

The coefficient of static friction between the shoes at A and B of the tongs and the pallet is μs1 andbetween the pallet and the floor μs2. If a horizontal towing force P is applied to the tongs,determine the largest mass that can be towed.

779




μs1 0.5= a 75 mm=

μs2 0.4= b 20 mm=

P 300 N= c 30 mm=

g 9.81m

s2= θ 60 deg=

Solution:

Assume that we are on the verge ofslipping at every surface.

Guesses

T 1N= NA 1N=

F 1N= Nground 1N=

FA 1N= mass 1kg=

Given

2 T sin θ( ) P− 0=

T− sin θ( ) b c+( ) T cos θ( ) a− FA b− NA a+ 0=

FA μs1 NA=

2 FA F− 0=

Nground mass g− 0=

F μs2 Nground=

T

NA

FA

F

Nground

mass

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

Find T NA, FA, F, Nground, mass,( )=

780

Given:




T

NA

FA

F

Nground

⎛⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎠

173.21

215.31

107.66

215.31

538.28

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

N= mass 54.9 kg=

Problem *8-20

The pipe is hoisted using the tongs. If the coefficient of static friction at A and B is μs, determine thesmallest dimension b so that any pipe of inner diameter d can be lifted.

Solution:

W 2 FB− 0=

W2

⎛⎜⎝

⎞⎟⎠

b NB h− FBd2

⎛⎜⎝

⎞⎟⎠

− 0=

Thus

FBW2

=

NBW 2 b d−( )

4h=

Require

FB μs NB≤W2

μs W 2b d−( )

4 h≤ 2 h μs 2b d−( )≤ b

hμs

d2

+>

Problem 8-21

A very thin bookmark having a width a. is in the middle of a dictionary of weight W. If thepages are b by c, determine the force P needed to start to pull the bookmark out.The coefficientof static friction between the bookmark and the paper is μs. Assume the pressure on each pageand the bookmark is uniform.

781




Given:

a 1 in=

W 10 lb=

b 8 in=

c 10 in=

μs 0.7=

Solution:

Pressure on book mark :

P12

Wb c

= P 0.06lb

in2=

Normal force on bookmark: N P c a=

F μs N= F 0.44 lb=

ΣFx = 0; P 2F− 0= P 2F= P 0.88 lb=

Problem 8-22

The uniform dresser has weight W and rests on a tile floor for which the coefficient of frictionis μs. If the man pushes on it in the direction θ, determine the smallest magnitude of force Fneeded to move the dresser. Also, if the man has a weight Wman,, determine the smallestcoefficient of static friction between his shoes and the floor so that he does not slip.

Given:

W 90 lb=

μs 0.25=

782




Wman 150 lb=

θ 0 deg=

Solution:

Dresser: Guesses ND 1lb= F 1lb=

Given

+↑Σ Fy = 0; ND W− F sin θ( )− 0=

+→ Σ Fx = 0; F cos θ( ) μs ND− 0=

ND

F

⎛⎜⎝

⎞⎟⎠

Find ND F,( )= F 22.50 lb=

Man: Guesses Nm 1lb= μm 0.2=

Given

+↑Nm Wman− F sin θ( )+ 0=Σ Fy = 0;

+→ Σ Fx = 0; F− cos θ( ) μm Nm+ 0=

Nm

μm

⎛⎜⎝

⎞⎟⎠

Find Nm μm,( )= μm 0.15=

Problem 8-23

The uniform dresser has weight W and rests on a tile floor for which the coefficient of frictionis μs. If the man pushes on it in the direction θ, determine the smallest magnitude of force Fneeded to move the dresser. Also, if the man has a weight Wman, determine the smallestcoefficient of static friction between his shoes and the floor so that he does not slip.

Given:

W 90 lb=

783




μs 0.25=

Wman 150 lb=

θ 30 deg=

Solution:

Dresser: Guesses ND 1lb= F 1lb=

Given

+↑Σ Fy = 0; ND W− F sin θ( )− 0=

+→ Σ Fx = 0; F cos θ( ) μs ND− 0=

ND

F

⎛⎜⎝

⎞⎟⎠

Find ND F,( )= F 30.36 lb=

Guesses Nm 1lb= μm 0.2=Man:

Given

+↑Nm Wman− F sin θ( )+ 0=Σ Fy = 0;

+→ Σ Fx = 0; F− cos θ( ) μm Nm+ 0=

Nm

μm

⎛⎜⎝

⎞⎟⎠

Find Nm μm,( )= μm 0.195=

Problem 8-24

The cam is subjected to a couple moment of M. Determine the minimum force P that should beapplied to the follower in order to hold the cam in the position shown.The coefficient of staticfriction between the cam and the follower is μs. The guide at A is smooth.

784




Given:

a 10 mm=

b 60 mm=

M 5 N m⋅=

μs 0.4=

Solution:

ΣM0 = 0; M μs NB b− a NB− 0=

NBM

μs b a+=

NB 147.06 N=

Follower:

ΣFy = 0; NB P− 0=

P NB=

P 147 N=

Problem 8-25

The board can be adjusted vertically by tilting it up and sliding the smooth pin A along the verticalguide G. When placed horizontally, the bottom C then bears along the edge of the guide, where thecoefficient of friction is μs. Determine the largest dimension d which will support any appliedforce F without causing the board to slip downward.

Given:

μs 0.4=

a 0.75 in=

b 6 in=

785




Solution:

+↑Σ Fy = 0;

μs NC F− 0=

ΣMA = 0;

F− b d NC+ μs NC a− 0=

Solving we find μs− b d+ μs a− 0= d μs a b+( )= d 2.70 in=

Problem 8-26

The homogeneous semicylinder has a mass m and mass center at G. Determine the largest angle θof the inclined plane upon which it rests so that it does not slip down the plane. The coefficient ofstatic friction between the plane and the cylinder is μs. Also, what is the angle φ for this case?

Given:

μs 0.3=

Solution:

The semicylinder is a two-force member:

Since F μ N=

tan θ( )μs N

N= μS=

θ atan μs( )=

786




θ 16.7 deg=

Law of sines

rsin 180 deg φ−( )

4r3π

sin θ( )=

φ asin3π4

sin θ( )⎛⎜⎝

⎞⎟⎠

=

φ 42.6 deg=

Problem 8-27

A chain having a length L and weight W rests on a street for which the coefficient of staticfriction is μs. If a crane is used to hoist the chain, determine the force P it applies to the chainif the length of chain remaining on the ground begins to slip when the horizontal component isPx. What length of chain remains on the ground?

Given:

L 20 ft=

W 8lbft

=

μs 0.2=

Px 10 lb=

Solution:

ΣFx = 0; Px− μs Nc+ 0=

NcPx

μs=

Nc 50.00 lb=

ΣFy = 0; Py W L− Nc+ 0=

787




Py W L Nc−=

Py 110.00 lb=

P Px2 Py

2+=

P 110 lb=

The length on the ground is supported by Nc 50.00 lb= thus

LNcW

=

L 6.25 ft=

Problem 8-28

The fork lift has a weight W1 and center of gravity at G. If the rear wheels are powered,whereas the front wheels are free to roll, determine the maximum number of crates, each ofweight W2 that the fork lift can push forward. The coefficient of static friction between thewheels and the ground is μs and between each crate and the ground is μ's.

Given:

W1 2400 lb=

W2 300 lb=

μs 0.4=

μ's 0.35=

a 2.5 ft=

b 1.25 ft=

c 3.50 ft=

Solution:

Fork lift:

ΣMB = 0; W1 c NA b c+( )− 0=

NA W1c

b c+⎛⎜⎝

⎞⎟⎠

= NA 1768.4 lb=

ΣFx = 0; μs NA P− 0=

788




P μs NA= P 707.37 lb=Crate:

Nc W2− 0=ΣFy = 0;

Nc W2= Nc 300.00 lb=

ΣFx = 0; P' μ's Nc− 0=

P' μ's Nc= P' 105.00 lb=

Thus nPP'

= n 6.74= n floor n( )= n 6.00=

Problem 8-29

The brake is to be designed to be self locking, that is, it will not rotate when no load P is appliedto it when the disk is subjected to a clockwise couple moment MO. Determine the distance d ofthe lever that will allow this to happen. The coefficient of static friction at B is μs.

Given:

a 1.5 ft=

b 1 ft=

μs 0.5=

Solution:

ΣM0 = 0; M0 μs NB b− 0=

NBM0

μs b=

ΣMA = 0; P 2 a NB a− μs NB d+ 0=

789




P 0=

daμs

=

d 3.00 ft=

Problem 8-30

The concrete pipe of weight W is being lowered from the truck bed when it is in the positionshown. If the coefficient of static friction at the points of support A and B is μs determinewhere it begins to slip first: at A or B, or both at A and B.

Given:

W 800 lb= a 30 in=

μs 0.4= b 18 in=

θ 30 deg= c 5 in=

r 15 in=

Solution:

initial guesses are

NA 10 lb= NB 10 lb= FA 10 lb= FB 10 lb=

Given Assume slipping at A:

ΣFx = 0; NA FB+ W sin θ( )− 0=

ΣFy = 0; FA NB+ W cos θ( )− 0=

ΣM0= 0; FB r FA r− 0=

FA μs NA=

NA

NB

FA

FB

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

Find NA NB, FA, FB,( )=

NA

NB

FA

FB

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

285.71

578.53

114.29

114.29

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

lb=

790



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