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```
The coeff of friction b/w the block A of mass m and triangular wedge B of mass M is @. There is no friction b/w the wedge and the plane. If the system (blockA + wedge B) is released so that there is no sliding b/w A and B, show that the max possible inclination theta is equals to tan -1 @. The coeff of friction b/w the block A of mass m and triangular wedge B of mass M is @. There is no friction b/w the wedge and the plane. If the system (blockA + wedge B) is released so that there is no sliding b/w A and B, show that the max possible inclination theta is equals to tan -1 @.

```
11 years ago

```							 hi,

if  there no frctn b/n B & PLANE
then, accelrn  of (A+B) wud be g sin(theta). along the plne indwnward diectn,

now cosider the A w.r.t B forces on A    1) psuedo f = m* gsin(theta) upward along th pln
2) gravity = mg vertical dwnwrd
3)max. friction force = @*{mg-mg sin^2(theta)}

mg sin^2(theta) is vertical compnent of psuedo f

{mg-mg sin^2(theta)} = total dwn force

horizontal comp of psuedo f = mg*sin(theta)* cos(theta)

now in limiting case max frctn f = mg cos^2(theta)

=>  @*mg {1- sin^2(theta)}=mg*sin(theta)*cos(theta)
=> @ = tan(theta)
=>  theta = tan^-1@.

```
11 years ago
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