Wedge And Block
A block goes on a wedge which in turns moves on a horizontal table, as given in figure. The wedge angle is Θ.
As long as the wedge is in met with table, we have the trivial constraints that the vertical acceleration of the wedge is zero. To search the constraint, let x be the horizontal co-ordinate of the end of the wedge and let y and x be the horizontal and vertical coordinate of the block as shown then
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(x -X) = (h - y) cot Θ
Even if the friction between the block and wedge affect their acceleration, above relation is remains as long as they are in contact.
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Pulley System
In this part we will consider all the pulley problems.
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In case of the entire pulley problem during entire motion string always fixed. By that property we will give results of following two types.
Case - I: "pulley is both of its last ends and fix are free to move"
Result
A1 = A2 = Relation because of string property.
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If we suppose that both blocks are coming down then
and -A1 = A2
So it is very important to notice that we can assume the acceleration to be in any direction but result will apply only considering it in opposite direction.
Case - II: "pulley and its sections are free to move"
A3 = (A1+A2)/2
Proof: Let us displace block1, x1 towards right, block 2, x2 towards left and
l1 + l2 + l3 + l4 = l1 - x1 + l2 - x1 + l3 + x3 + l3 + x3
(Length of string will not change)
(X1+X2)/2=X3 
(A1+A2)/2=A3
APPLICATIONS INVOLVING PULLEY
Frame the equations to find the acceleration of all the blocks.
Solution: First write the tension on each string and assume the acceleration of all blocks in any direction and divide pulley in cases
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Equation for Newton's law for m1
m1g - T = m1A1 ... (1)
Equation for Newton's law for m2
m2g - T = m2A2 ... (2)
Equation for Newton's law for m3
m3g - T = m3A3 ... (3)
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Now apply case - I and case II in pulley (i) and pulley (ii)
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A1=-(A2+A3)/2
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