solved examples sliding friction

Worksheet Sliding Friction 1

Question1. A horizontal force of 400.0 N is required to pull a 1760 N

trunk across the floor at constant speed. Find the

coefficient of sliding friction.

1760 N

400.0 N

FN

FF What are you to find?

What do you know?

Draw a free body diagram.

maF

What equation uses this variable?

What equation will enable you to find Ff?

N

f

F

F

000.400 FF

0.400FF

What equation will enable you to find FN?

maF

01760NF

1760NF

1760

0.400227.

Question2. How much force must be applied to push a 1.35 kg book

across the desk at constant speed if the coefficient of

sliding friction is 0.30?

What are you to find?

What do you know?


kgm 35.1

maF



AF

0FA FF

AF FF


Nf FF

))(30(. Nf FF0WN FF

30.

FW

FN

FAFF

maF

WN FF

What equation will enable you to find FW?

mgFW

)81.9)(35.1(WF

NFW 2.13)2.13)(30(.fF

NF f 0.4 N0.4

Question3. A force of 105 N is applied horizontally to a 20.0 kg box to

move it across a horizontal floor. If the box has an

acceleration of 3.0 m/s2, find the coefficient of sliding friction.


What do you know?


kgm 0.20

maF



maFF105

45)0.60105(FF


0WN FF

2/0.3 sma

FW

FN

105 NFF

maF

WN FF


mgFW

)81.9)(0.20(WF

NFW 196

N

f

F

F

)0.3)(0.20(105 FF

196

4523.

Question4. A 1500.0 N force is exerted on a 200.0 kg crate to move it

across the floor. If the coefficient of sliding friction is

0.250, what is the crate’s acceleration?


What do you know?


kgm 0.200

maF


a

aF f )0.200(0.1500


WN FF

250.0

FW

FN

1500.0 NFF

Nf FF

Wf FF )250.0(


mgFW

)81.9)(0.200(WF

NFW 1960)1960)(250.0(fF

NFf .490

a)0.200(.4900.1500a)0.200(.1010

asm 2/050.5

Question5. A 100.0 kg commuter is standing on a train accelerating at 3.70 m/s2.

what coefficient of static friction must exist between the commuter’s

feet and the floor to avoid sliding?

kgm 0.100


What do you know?


maF


maFF f


WN FF

2/70.3 smaFW

FN

FF

Nf FF

)7.980(.370


mgFW

)80665.9)(0.100(WF

NFW 7.980

)70.3)(0.100(fF

NFf .370

NFN 7.980

377.0

Question6. A 146 N force is used to pull a 350. N wood block at constant speed

by a rope making an angle of 50.0° with the floor. Find the

coefficient of sliding friction.

NFW .350


What do you know?


maF


0AXf FF


maF

NFA 146

Nf FF

)238(8.93


0NAYW FFF

)0.50sin(146AYF

0112.350 NF

)0.50cos(146AXF

08.93fF

NFN 238

394.0

FW

FNFA

50.0°FfFAX

FAY

N8.93 N112

NFf 8.93

Question7. A 75.0 kg baby carriage is pushed along a level sidewalk by exerting a force

of 50.0 N on the handle, which makes an angle of 60.0° with the horizontal.

What is the coefficient of friction between the carriage and the sidewalk?

kgm 0.75


What do you know?


maF


0AXf FF


maF

NFA 0.50

Nf FF

)779(0.25What equation will enable you to find FN?

0NAYW FFF

)0.60sin(0.50AYF

03.43736 NF

)0.60cos(0.50AXF

00.25fF

NFN 779

321.0

FA

FW

FN

60.0°

Ff FAX

FAY

N0.25

N3.43NFf 0.25

)81.9)(0.75(mgFW

N736

Question8. A 3.00 kg wood box slides from rest down a 35.0° inclined plane.

How long does it take the box to reach the bottom of the 4.75 m

wood incline? (coefficient of friction = 0.30)

kgm 00.3


What do you know?


maF


tmaFF WXf

What equation will enable you to find a?

30.0

O

O

x

x

t

a

v

v

Nf FF

0.35cosWWY FFWYN FF

)81.9)(00.3(mgFW

35sin4.29WXFN9.16

35.0°

0

75.4

?

0

m

2

21 attvxx OO

FN

Ff

FW

FWX

FWY

NFW 4.29

N1.24

)1.24)(30.0( N23.7

a00.39.1623.7 a 2/22.3 sm 2/22.3 sm

Question8. A 3.00 kg wood box slides from rest down a 35.0° inclined plane.

How long does it take the box to reach the bottom of the 4.75 m

wood incline? (coefficient of friction = 0.30)

kgm 00.3


What do you know?



t

What equation will enable you to find a?

30.0

O

O

x

x

t

a

v

v

35.0°

0

75.4

?

0

m

2

21 attvxx OO

FN

Ff

FW

FWX

FWY

2/22.3 sm2

21 )22.3(0075.4 tt

22.3

)75.4(2t sec72.1

Question9. A 65.0 kg crate is to be accelerated at 7.00 m/s2 up an incline making an

angle of 25.0° angle with the horizontal. If the coefficient of sliding friction

between the crate and the incline is 0.200, how much force is required?

kgm 0.65


What do you know?



AF

2/00.7 sma

25.0°

FN

FW

FWX

FWY

Ff

FA

200.0

maF

maFFF WXfA

)81.9)(0.65(mgFW

N6380.25sin638WXF

Nf FF

WYN FF

0.25cos638WYF

NFWY 578

NFf 116)578)(200(.

)00.7)(0.65(.270116AFNFWX .270

NFA 841

Question10. A 60.0 kg crate is attached to a weight by a cord that passes over a frictionless pulley,

as shown in the diagram. (a) If the coefficient of friction is 0.500, what weight will keep

the crate moving up the 40.0° incline at a constant speed? (b) If the cord is cut when

the crate is at rest at the top of the incline, how far would the crate have slid by the

time its speed reached 7.50 m/s?

kgm 0.60


What do you know?



WF

40.0°

FN

FW1

FW1X

FW1Y

500.0

maF

wAXWfA FFandmaFFF 1

NmgFW 589)81.9)(0.60(1

0.40sin5891XWF

2/0 sma

Ff

FA

Fw

FA

NF XW 3791

Nf FF

YWN FF 1

0.40cos5891YWF

NF YW 4511

N226)451)(500.0(maFFF XWfA 1

0379226AF

NFA 605

Question10. A 60.0 kg crate is attached to a weight by a cord that passes over a frictionless pulley,

as shown in the diagram. (a) If the coefficient of friction is 0.500, what weight will keep

the crate moving up the 40.0° incline at a constant speed? (b) If the cord is cut when

the crate is at rest at the top of the incline, how far would the crate have slid by the

time its speed reached 7.50 m/s?

kgm 0.60


What do you know?



x

40.0°

FN

FW1

FW1X

FW1Y

500.0 )(222

OO xxavvmaF

Ff

FA disappears and Ff is now up the incline

0

0

/50.7

O

O

x

x

t

a

v

smv

You need to find a

maFF fXW1

You know FW1X and Ff

from part A

2/55.2 sma

xa

vv O

2

22

xm0.11

x)55.2(2

0)50.7( 22

solved examples sliding friction

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