PROBLEMS (HARD TYPE)
1) Two blocks A and B of mass 2kg and 4kg are placed one over the other as shown in figure (41). A time varying horizontal force F = 2t is applied on the upper block as shown in figure. Here t is in second and F is in Newton . Draw the graph showing acceleration of A and B on y-axis and time on x-axis. Coefficient of friction between A and B is m = ½ and the horizontal surface over which B is placed is smooth. (g=10m/s2).
Fig (41)
Solution:
Limiting friction between A and B is
Block B moves due to friction only, therefore maximum acceleration of B can be
Thus both the blocks move together with same acceleration till the common acceleration becomes 2.5 m/s2, after that acceleration of B will become constant while that of A will go on increasing. To find the time when the acceleration of both the blocks becomes 2.5m/s2 (or when slipping will start between A and B) we will write
t = 7.5 Sec
Hence for t £ 7.5 S
Thus aA versus t or aB versus t graph is a straight line passing through origin of slope 1/3.
For t ³ 7.5S
aB = 2.5 m/s2 = constant.
And 
Or 
Or aA = t-5
aA versus t graph is a straight line of slope 1 and intercept -5 while aB versus t graph is a straight line parallel to x-axis the corresponding graph is shown in figure.
Dumb Question:
1) Explain the graph?
Ans: The force is being acted upon the upper block. Now obviously the friction retards the upper block while it accelerates the lower one. Now when the limiting friction is reached, the upper just moves and its acceleration become different from lower block, this happens at 2.5 m/s2. Now the friction becomes constant and so the lower block moves with constant acceleration 2.5 m/s2 while the upper accelerates further as 
increases rapidly with time.
2) Block A of mass m and block B of mass 2m are placed on a fixed triangular wedge by means of a light and inextensible string and a frictionless pulley as shown in figure (42) the wedge is inclined at 450 to the horizontal on both sides the coefficient of friction between the block A and wedge is 2/3 and that between the block B and the wedge is 1/3. If the system of A and B is released from rest, then find,
A) The acceleration of A
B) Tension in the string
C) The magnitude and direction of the frictional force acting on A.
Fig (42)
Solution:
A) In absence of friction the block B will move down the plane and the block A will move up the plane. Frictional force opposes this motion.
T-mgSin450 –f1 = ma ---------------- (1)
And 2mgSin450 – f2 – T =2ma ---------------- (2)
From (1) and (2)
mgSin450 – (f1+f2) = 3ma
For ‘a’ to be non-zero mgSin450 must be greater than maximum value of (f1+f2).
Þ mgSin450 < (f1 +f2) max
Hence blocks will remain stationary.
B) F.B.D of the block B.
But 
therefore block B has tendency to slide down the plane.
For block B to be at rest
C)
T (tension) is greater than mgCos450, Hence block A has tendency to move up the plane, therefore frictional force on the block A will be down the plane.
From F.B.D of A, For A to be at rest
3) A small body is placed on the top of a smooth hemisphere of radius R. The hemisphere is imparted a constant acceleration a0 in the horizontal direction and the body starts down Find,
A) Velocity of the body relative to the sphere at the moment of break off.
B) The angle q0 between the vertical and the radius vector drawn from the center of the sphere to the break off point. Calculate q0 for a0 = g.
Fig (43)
Solution:
At an arbitrary moment when body is at an angle q with vertical. Rewriting the second law of motion in projection forms:
At the break of point N =0, q = q0 and let v=v0 so equation (1) becomes
Solving equation (2) and (3) we get
Dumb Question:
1) How is 
Ans:
Here, ds = Rdq
So, 
KEY WORDS
Ø Forces
Ø Law’s of Motion
· 1st Law
· 2nd Law
· 3rd Law
Ø Free Body Diagram
Ø Constraint Motion
Ø Spring Force
Ø Friction
· Static Friction
· Kinetic Friction
· Belt Friction
Ø Pseudo Force
