string and the pulley are light. The blocks are moving with constant speed. Then find the value of F and T₁.

                      Fig (27)

Solution:

For block 2m T₁-f₁ = 0

Þ T₁ - m (2mg) = 0

Þ T₁ = 2mmg

For block 3m

f₁ + T₁+ f₂- F = 0

Þ F = 2mmg+2mmg+5mmg

Þ F = 9mmg.

6) Find the maximum value of M/m in situation shown in figure (28). The system remains in rest. Friction coefficient at both the contact is m. Discuss the situation when tanq < m.

                    Fig (28)

Solution:

The equilibrium is limiting. Hence frictional forces are equal to m times the corresponding normal forces.

Equilibrium of the block m gives

T = mN₁ and N₁ = mg

Which gives T = mmg                                                                   -------------- (1)

Next consider the equilibrium of the block M. Taking components parallel to the incline

T + mN₂ = MgSinq.

Taking components normal to the incline

N₂ = mgCosq

These give T = Mg (Sinq - mCosq)                                            --------------- (2)

From (1) and (2) mmg = Mg (Sinq - mCosq)

If tanq <m, (Sinq - mCosq) <0 and system will not slide for any value of M/m.

7) A small body was launched up an incline plane with angle a = 15⁰ against the horizontal. Find the coefficient friction. If the time of the ascent of the body is n=2 times less than the time of descent.

                  Fig (29)

Solution:

Step 1: When the body is moving up on the inclined plane

Here N = mgCosa

The acceleration of the body is

Let body is projected with speed V₀ along the inclined plane (along the x-axis). After time t, body reaches at point P (V =0) as shown.

Let OP = S

Step 2: When body starts to return from point P the acceleration is,

Putting the value of t₁ and t₂ from equation (1) and (2) we get

8) Find the acceleration of w of body 2 as shown in figure (30). If it’s mass is m times as great as the mass of bar 1 and the angle that the inclined plane forms with the horizontal is equal to a the masses of the pulleys and the threads as well as the friction are assumed to be negligible. Look into possible cases.

                                                            Fig (30)

Solution:

Force diagram for separate body is as shown in figure.

From Fig (A)          T₁-m₁gSina = m₁w₁                                             --------------- (1)

From Fig (B)           T₁ = 2T₂                                                              --------------- (2)

Form Fig (C)           m₂g – T₂ = m₂w₂                                                 --------------- (3)

Since total work done by the force tension on a system is zero. Let in small time t, m₁ shifts upward for small distance x₁ and m₂ shifts downward for small distance x₂.

åw_T = 0 or T₁x₁ – T₂x₂ = 0

Or 2T₂x₁ – T₂x₂ = 0          x₂ = 2x₁.

                                                                                  -------------- (4)

w₂ = 2w₁                                                                                        -------------- (5)

According to problem m₂ = hm₁

Solving equation (1), (2), (3), (4) and (5) we get

But  are in the same direction.

9) In figure all the surfaces are assumed to be frictionless. The block of mass m on the prism which in turn slides backward on the horizontal surface. Find the acceleration of the smaller block with respect to the prism?

               Fig (31)

Solution:

Let the acceleration of prism be a₀ in the backward direction. Consider the motion of the smaller block from the frame of the prism.

Forces on the block are

1)      N normal force

2)      mg downward (gravity)

3)      ma₀ forward (pseudo)

The block slides down the plane. Components of the forces parallel to the incline give

ma₀Cosq + mgSinq = ma

Or a = a₀Cosq + gSinq                                                                 ----------------- (1)

Components of the forces perpendicular to the incline give –

N + ma₀Sinq = mgCosq                                                             ------------------- (2)

Now consider the motion of the prism from the lab frame. No pseudo force is needed as the frame used is inertial.

The forces are

1)      mg downward

2)      N normal to the incline ( by the block)

3)      N¹ upward ( by the horizontal surface)

Horizontal components give

Putting in equation (2)

PROBLEMS (MEDIUM TYPE)

1) Using Constraint equations find the relation between a₁ and a₂?

  Fig (32)

Solution:

Points 1, 2, 3, 4 are movable. Let their displacements from a fixed line be x₁, x₂, x₃ and x₄.

On double differentiating w.r.t time we get,

Solving (1), (2) and (3) we get,

Desired relation between a₁ and a₂