FRICTION
BLOCK FRICTION
TEST
TEST :
DURATION : 1HR
MARKS: 30
(1) Two blocks A (100 N) and B (150 N) are resting on the ground as
shown in figure. Find the minimum value of weight P in the pan so that
the motion starts. Find whether B is stationary with respect to ground
and A moves or B is stationary with respect to A.
(2)
A uniform rod AB of length 10 m and weight 280 N is hinged at B and
end A rests on a block weighing 400N. If = 0.4 for all contact
surfaces, find horizontal force P required to start moving 400 N block.
(3)
A 60 kg cabinet is as shown. Coefficient of static friction between
cabinet and floor is 0.35. Determine (a) The force P required to move
the cabinet to the right (b) The largest allowable value of h if the cabinet
is not to tip over.
FRICTION
BLOCK FRICTION
TEST
Solutions :
(1)
For finding the minimum value of P, let us analyze two separate cases
first
Case (i) : B is stationary with respect to ground and A moves.
F.B.D. of system
{Since the force T will try to move block A to right, direction of frictional force
will be to the left on block A and hence will to the right for block B.
Also since after applying weights in pan the motion is going to be to the
right hence direction of friction between block B and floor will be to the left.}
For Block A,
Fx = 0 ( positive)
T cos 30 0.3 N1 = 0
Fy = 0 ( positive)
N1 + T sin 30 100 = 0
T = 29.52 N and N1 = 85.23 N
For Block B,
Fy = 0 ( positive)
N1 + T sin 30 100 = 0
N2 N1 150 = 0
N2 = 235.23 N
{Since motion happens along x-axis
Normal reaction can be found out
using Fy = 0}
2
FRICTION
BLOCK FRICTION
TEST
Fx = 0.3 N1 0.1 N2 Pls see this step
= 2.05 N ()
Hence there is a resultant force acting on block B. Therefore block B
will move to the right and not remain stationary.
Case (ii) : A and B move together i.e. block B is stationary with respect to A.
F.B.D. of system
Fx = 0 ( positive)
T cos 30 0.1 N1 = 0
Fy = 0 ( positive)
T sin 30 + N1 250 = 0
T = 27.29 N and N1 = 236.35 N
Here as we see amongst the two cases the value of T or weight in pan
is less for Case (ii), i.e. T = 27.29 N.
In fact after this the lower block will be set into motion as seen from
case (i).
Pmin = 27.29 N
{ Tension in rope = weight in pan}
B is stationary with respect to A
(2) Solution :
{When we apply force P on block of 400N, point A which is in contact
with the block will also cause a frictional force to develop}
F.B.D. of rod AB
FRICTION
BLOCK FRICTION
TEST
{If we apply force P on 400N block, the frictional force on block will be
to the left at point A.
Hence due to Newtons third law an equal and opposite frictional force
of 0.4 N1 will be given by block on rod}
MB = 0 ( positive)
N1 10 cos 36.86 + 0.4 N1 10 sin 36.86
+ 280 5 cos 36.86 = 0
N1 = 200 N
{The distances for moment are as shown below
F.B.D. of block of 400 N
Fx = 0 ( positive)
P 0.4 N1 0.4 N2 = 0
P 0.4 N1 = 80
Fy = 0 ( positive)
N2 N1 400 = 0
N2 = 600 N
Substitute above in equation (1)
P = 320 N
(1) { N1 = 200 N}
FRICTION
(3)
BLOCK FRICTION
TEST
Solutions:
(a) Cabinet impends (slips) to right
For finding force P to move the cabinet to right we draw the F.B.D. of
cabinet as shown.
Fy = 0 ( positive)
N1 = 60 9.81
N2 = 588.6 N
Fx = 0 ( positive)
P 0.35 N1 = 0
P = 206 N
(b)
{Now in the second part, the question says that the body should not tip
over i.e. we will take the tipping condition here to find value of h}
F.B.D. of block
Consider point A to be the corner of the cabinet.
At the tipping instant the point A will be in contact to the ground.
MA = 0 (
positive)
P h + 60 9.81 0.25 = 0 { weight acts at the centre}
Therefore h= 0.714 m
Note: we took the same value for P i.e. 206 N.
