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Engineering Mechanics

Engineering Mechanics

L14: Friction: Applications

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Indian Institute of Technology Jodhpur

Engineering Mechanics

A wedge is one of the simplest and most useful machines. A wedge is used to produce small adjustments in the position of a body or to apply large forces.Wedges largely depend on friction to function. When

Wedge: Raising Weight

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m

Pα

R3

mg

R3

mgR2

q+a

R2f+a

fR1

fR1

fR2

f R2

P

P

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Engineering Mechanics

Force P = 0 (R1 = R2 for equilibrium)

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________x

y

R1

R2

a

af

f-a

f

f

a

2f-a

2f-a

f-a

f-af

f-a

fR2

R1

y

a

Slipping impendingat upper surface

Slipping impendingat lower surface

Range of R1 = R2

for no slip

Engineering Mechanics

Wedge: lowering weight

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mgR3

R2

R2R2

R3

R2

R1

R1

P

mg

f-a

f-a

f

f

f

f

m

Pα

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Engineering Mechanics

The horizontal position of the 500-kg rectangular block of concrete is adjusted by the 5o wedge under the action of the force P. If the coefficient of static friction for both wedge surfaces is 0.30 and if the coefficient of static friction between the block and the horizontal surface is 0.60, determine the least force P required to move the block.

Example 2

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P

500kg50

Engineering Mechanics

Algebraic solution

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P

500kg50

500(9.81)N

P

f2=tan-10.06=31.00

f1=tan-10.03=16.700

R1f1

R3f2

R2

f1

-R2f1α

a

a

f2b

b α +f1

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Engineering Mechanics

Graphical Solution

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P

500kg50

500(9.81)N

P

f2=tan-10.06=31.00

f1=tan-10.03=16.700

R1f1

R3f2

R2

f1

-R2f1α

a

a

f2b

b α +f1

PC

B

W=4905N

A

-R216.700

R2

16.700

R3 31.00

R116.700+50=21.70

Block

Wedge

Engineering Mechanics

The impending slippage of flexible cables, belts, and ropes over sheaves and drums is important in the design of belt drives of all types, band brakes, and hoisting rigs.Drum is subjected to the two belt tensions T1 and T2, the torque M necessary to prevent rotation, and a bearing reaction R. With M in the direction shown, T2is greater than T1. The

FLEXIBLE BELTS

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r

T1

T2

R

M

r

qdq

b

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Engineering Mechanics

FLEXIBLE BELTS

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r

T1

T2

R

M

qdq

bt

rdN

dq

n

T+dT

!"2

µdN

T

!"2

Engineering Mechanics

FLEXIBLE BELTS

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t

rdN

dq

n

T+dT

!"2

µdN

T

!"2

$cos !"2 + )!* = $ + !$ cos !"2

)!* = !$

!* = $ + !$ sin ./0 + $ sin./0

!* = $!"

!$$ = )!"

123

24 !$$ = 1

5

6)!"

ln $0$8= )9

$0 = $8:;6

t: n:

sin !"2 = !"2cos !"2 = 1

Combining both

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Engineering Mechanics

Deformation at the point of contact between a rolling wheel and its supporting surface introduces a resistance to rolling, This resistance is not due to tangential friction forces Therefore is an entirely different phenomenon from that of dry friction.

ROLLING RESISTANCE

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Engineering Mechanics

The resultant R of this distribution acts at some point A and must pass through the wheel center for the wheel to be in equilibrium.

We find the force P necessary to maintain rolling at constant speed by equating the moments of all forces about A to zero.

This gives us

ROLLING RESISTANCE

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! = #$ % = &'%

&' =#$

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Engineering Mechanics

Determine the range of mass m over which the system is in static equilibrium. The coefficient of static friction between the cord and the upper curved surface is 0.20, while that between the block and the incline is 0.40. Neglect friction at the pivot O.

Example 1

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Engineering Mechanics

Motion of m impends up the incline.

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AG O

OX

OYL/2

L/69(9.81)N

250

TA =T2350

y

T1mg

x

400

N 0.4N

= T2

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Engineering Mechanics

Motion of m impends down the incline

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AG O

OX

OYL/2

L/69(9.81)N

250

TA =T1350 mg

N

0.4N

400x

T2

= T1

16/09/19

Thanking you16

References :