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Wedge Friction

A block of small piece with two of their opposite faces not parallel as shown in Fig. 5.6 is called wedge. It is used to lift another block of weight W by applying a horizontal force P. The reaction on block W is perpendicular to meeting surface and it is always greater than the total downward force applied by the block W. The net effect will cause a resultant force acting in a upward direction on the block W and will move it up.

SOLVED EXAMPLES

 

Example 5.1. The force Pis applied to a block of mass 9 Kg, which is stationary before the force applied. Determine the magnitude and direction of the friction force F exerted by the horizontal surface on the block if (a) P = 30 N (b) P = 40 N (c) P =50 N. (Re( Fig. 5.l(a)).

Solution.

µs N 1 – mg sin α — P = 0

Ʃ FY = 0

N1 –mg  cos α = 0

 

From  (1) =>                                       p= µS  N1 – mg sin α

= µS (mg cos α )  -mg sin α

 

 

Condition Just Before Tipping

Taking moment about A

 

Ʃ MA = 0

P x d  +  mg sin α x d /2 – mg cos α x w /2

 

Substituting the value of P from (3), we get

Mgd (µS cos α – sin α) + mgd . sin α /2  = mg cos α x w /2

Or                                   d (µS cos α –sin  α + sin α / 2)  = cos α x w/2

Or                                      d (µS cos α –sin  α + sin α / 2)  = w . cos α /2

Or                                                       

Or                                                        d/w  = 1 / 2µ =tan α . proved

Example 5.3. Find the least force required to drag a body of weight W, placed on a  = Wrought inclined plane having inclination a to the horizontal. The force is applied to the body in such1 a way that it makes an angle  θ to the inclined plane and the body is (a) on  the point of  motion up the plane (b) on he point of ‘motion down the plane.

Ʃ FX =0

P =cos θ – W sin  α – µ N =0

P cos θ =W sin α + µN

      Ʃ FY =0

N +P sin θ – W cos α = 0

N = W cos α – P sin θ

 

Substituting the value of N in (1), we get

P cos θ = W sin α   +µ   (w cos α  – p sin θ  )

P cos θ = W sin α   +µ w cos α  – p sin θ

 

P (cos θ = µ sin θ)  + W   (sin α + µ cos α)

P = W  (sin α + µ cos α) / cos θ + µ sin θ

 

=W (sin α + tan ϕ cos α )  / (cos θ + tan ϕ sin θ)

= W (sin α cos ϕ + cos α sin ϕ ) / (cos θ cos ϕ + sin ϕ )

= W sin (α + ϕ) / cos (θ – ϕ)

 

P will be minimum when cos (θ – ϕ) is maximum i.e.; cos (θ -ϕ) == 1

i.e.,                                          θ =ϕ,

p min = W sin (α +ϕ) .

           Ʃ FX =0

P cos  θ + µN – W sin α = 0

W sin α = p cos θ + µN

 

Substituting the value of N in equation (3), we get

W  sin α = p cos  θ + µ (w cost α – p sin θ)

 

W sin α – µW  cos  α = p cos θ  - µp sin  θ

 

= W (sin α – µ cos  α) = p (cos θ – µ sin θ) / cos θ – tan ϕ sin θ

= W  (sin α cos ϕ – cos α sin ϕ ) / (cos θ cos ϕ – sin θ sin ϕ)

 

P = W sin (α –ϕ) / cos (θ + ϕ)

 

Example 5.4. A cord connects two bodies of weights 400 N and BOON. The two bodies are placed on an inclined plane and cord is parallel to inclined plane. The coefficients of friction for the weight of 400 N is 0.15 and that for BOO N is 0.4. Determine the inclination of the plane to the horizontal and the tension in the cord when the motion is about to take place, down the inclined plane. The body weighing 400 N is below down the body weighing BOO N.

Solution .

 

For 400 N body :

Ʃ FX =0

µ1 N1 + T – W1 sin α = 0

W 1  sin  α = T  + µ1  N1

400 sin  α = T + 0.15 N1

Ʃ FY =0

N1 – 400 cos α =0

N1 = 400 cos  α

 

Substituting N1 in (1)

(1) => 400 sin  α = T  + 0 .15 x 400 cos α

T = 400 sin α – 60 cos α

 

For 800 N body

Ʃ FX = 0

µ2 N2 – 800 sin α – T = 0

µ2 N2 = 800 sin α + T

0.4 N2 = 800 sin α + T

Ʃ FY =0

N 2 – 800 cos α = 0

N 2 = 800 cos α

 

Substituting the value of N2 from (5) in ( 4), we get

 

0.4  x 800 cos α = 800  sin  α +T

320 cos  α = 800 sin α + T

 

Substituting the value of T from (3) to (6), we get

320 cos α = 800 sin α + 400 sin α – 60 cos α

380 cos α = 1200 sin α

Tan α = 380 / 1200 = 0.3167

Α = 17.57 0.

From               T = 400 sin 17 .570 – 60 cos 17.570

= 63 .55 N

 

Example 5.5. Two blocks A and Bare connected by a horizontal rod and are supported on two rough planes as shown in Fig. P-5.5. If the weigh t of block B is 1500 N and coefficient of friction of block A and B are 0.25 and 0.35 respectively, find the smallest weight of block A for which equilibrium exists.

 

For Block A,

Ʃ FY =0

NA = WA

Ʃ FX = 0

T +µA NA = 0

T = – µA NA

= – 0.25  x WA = – 0.25WA

For Block B

Ʃ FY =0

N B –W B cos 60° + T sin 60° = 0

Ʃ FY =0

 

-WB sin 60 + µB NB – T cos 600 = 0

 

- 1500 sin 600 + 0.35 NB – T x ½ = 0

 

-1500 sin 600 + 0.35 (WB cos 600 – T sin 600 = T /2

 

[Substituting NB from (3)]

 

1500 sin 600 + 0.35 x 0.35 x 1500 cos 600 -0.35 x T sin 600 = ½

 

[Using eqn ( 1) and (2)]

 

- 1036.54 = 0.80 T,                                         T = – 1295.7.

 

WA = – T / 0.25 = + 1295 .7 / 0.25 = 5182.8 N