2) Two blocks A and B of mass 1kg and 2kg respectively are connected by a string, passing over a light frictionless pulley. Both the blocks are resting on a horizontal floor and the pulley is held such that string remains just taunt.
At moment t=0, a force F = 20tN starts acting on the pulley along vertically upward direction as in figure (33). Calculate
A) Velocity of A when B loses contact with the floor.
B) Height raised by the pulley up to that instant (g=10m/s2).
Fig (34)
Solution:
A) Let T be tension in the string then,
2T = 20t or T = 10t N.
Let the block A loses its contact with the floor at time t = t1 this happens when the tension in string becomes equal to the weight of A thus,
T = mg or 10t1 = 1´10 or t1 = 1 Sec ------------------ (1)
Similarly for block B we have,
10t2 = 2´10 or t2 = 2 Sec ------------------ (2)
That is the block B loses contact after 2 seconds. For block A, at time t such that t ³ t1. Let a be its acceleration in upward direction then,
On integrating,
Substituting 
We get v = 20-20+5 = 5 m/s ------------ (5)
B) From equation (5) dy = (5t2-10t+5) dt.
Where y is the vertical displacement of block A at time t (³ t1)
Integrating we have,
Height raised by pulley up to that instant = 
3) Figure shows two blocks connected by a light string placed on the two inclined parts of a triangular structure. The coefficients of static and Kinetic frictions are 0.28 and 0.25 respectively at each of the surfaces.
A) Find the maximum and minimum values of m for which the system remains at rest.
B) Find the acceleration of either block if m is given the minimum value calculates in the first part and is gently pushed up the incline for a short while.
Fig (35)
Solution:
A)
Taking the 2kg block as the system the forces on this block are shown in fig (a) with M = 2kg. It is assumed that m has its minimum value so that the 2kg block has a tendency to slip down. As the block is in equilibrium, the resultant force should be zero.
Fig (36)
Taking components perpendicular to incline.
Taking components parallel to the incline,
As it is a case of limiting equilibrium,
Now consider other block as the system. The forces acting on block are as shown in fig (b).
Taking component perpendicular to the incline
Taking components parallel to the incline,
As it is the case of limiting equilibrium,
From (1) and (2)
The maximum value of m can be obtained from (3) by putting ms = -0.28 thus maximum value of m is
B) If m = 9/8kg and the system is gently pushed, kinetic friction will operate thus,
Where ms = 0.25 if the acceleration is a, Newton ’s Second law for M gives in fig (a)
MgSin450 –T-f = Ma
Applying Newton ’s Second law for fig (b)
Dumb Question:
1) Why can the maximum value of ‘m’ be obtained by putting m = -0.28?
Ans: When the value of ‘m’ is maximum then ‘m’ slips down while 2kg block moves up. So the direction of the limiting friction simply reverses. And hence the maximum value can be obtained by just replacing ms by -ms.
4) Three blocks of masses m1, m2, m3 are connected as shown in figure (37). All the surfaces are frictionless and the string and the pulleys are light. Find the acceleration of m1.
Fig (37)
Solution:
Suppose acceleration of m1 is a0 towards right that will also be the downward acceleration of the pulley B because the string connecting m1 and B is constant in length. Also the string connecting m2 and m3 has a constant length. This implies that the decrease in the separation between m2 and B the increase in the separation between m3 and B. So the upward acceleration of m2 with respect to B equals the downward acceleration of m3 with respect to B. Let this acceleration be a.
The acceleration of m2 with respect to the ground = a0-a (downward) and the acceleration of m3 w.r.t the ground = a0+a (downward).
The acceleration will be used in Newton ’s Laws. Let the tension be T in the upper string and T1 in the lower string consider motion of pulley B.
The forces on this light pulley are:
1) T upwards by the upper string and
2) 2T1 downwards by the lower string
As the mass of the pulley is negligible,
2T1-T = 0
Giving T1 = T/2 --------------- (1)
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