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Course by: Sunil Kumar Singh. E-mail the author

# Friction between horizontal surfaces (application)

Module by: Sunil Kumar Singh. E-mail the author

Summary: Solving problems is an essential part of the understanding process.

Questions and their answers are presented here in the module text format as if it were an extension of the treatment of the topic. The idea is to provide a verbose explanation, detailing the application of theory. Solution presented is, therefore, treated as the part of the understanding process – not merely a Q/A session. The emphasis is to enforce ideas and concepts, which can not be completely absorbed unless they are put to real time situation.

## Representative problems and their solutions

We discuss problems, which highlight certain aspects of the study leading to friction between horizontal surfaces. The questions are categorized in terms of the characterizing features of the subject matter :

• An external force on the block
• An external force on the plank
• One external force each on block and plank
• Block with initial velocity

## An external force on the block

Problem 1 : In the arrangement shown in the figure, the mass of the pulleys and strings are negligible. The block “B” rests on the smooth horizontal surface, whereas the coefficient of friction between two blocks is 0.5. If masses of the blocks “A” and “B” are 2 kg and 4 kg respectively, then find the maximum force for which there is no relative motion between two blocks.

Solution : According to the question, two blocks have no relative motion between them. The friction force is less than limiting friction. As the limiting case, the maximum friction between the blocks is equal to limiting friction :

F S = μ N = μ m A g = 0.5 X 2 X 10 = 10 N F S = μ N = μ m A g = 0.5 X 2 X 10 = 10 N

Since blocks have no relative motion between them, they move as a single unit with a common acceleration, say “a”. The vertical and horizontal forces by the two brackets holding pulleys balance each other. Hence, these forces are not considered. The free body diagram of the combined body is shown on the left side of the figure. The friction forces between the blocks are internal forces for the combined body and as such not considered. Note that we have not drawn components of forces in vertical direction as the analysis in vertical direction is not relevant here.

a = F m A + m B = F 6 a = F m A + m B = F 6

F = 6 a F = 6 a

Considering free body diagram of block “B” as shown on the right side of the figure, we have :

a = F S m = 10 4 = 2.5 m / s 2 a = F S m = 10 4 = 2.5 m / s 2

Putting in the expression of force, we have,

F = 6 X 2.5 = 15 N F = 6 X 2.5 = 15 N

## An external force on the plank

Problem 2 : In the arrangement shown in the figure, mass of block and plank are 2 kg and 3 kg respectively. The coefficient of friction between block and plank is 0.5, whereas there is no friction between plank and underneath horizontal surface. An external force of 50 N is applied on the plank as shown. If the length of plank is 5 m, find the time after which block and plank are separated.

Solution : First, we need to know the nature of the friction between block and plank. If the friction is less than liming friction, then two entities will move together with an acceleration given by :

a = 50 2 + 3 = 10 m / s 2 a = 50 2 + 3 = 10 m / s 2

Let us use subscript “1” and “2” to denote block and plank respectively. Since friction is the only force on the block (1), the friction is

F 1 = m 1 a = 2 X 10 = 20 N F 1 = m 1 a = 2 X 10 = 20 N

The limiting friction, however, is :

F S = μ m g = 0.5 X 2 X 10 = 10 N F S = μ m g = 0.5 X 2 X 10 = 10 N

It means that our assumption that they move together is wrong. They move with relative velocity between them. The friction, therefore, is equal to kinetic friction and is nearly equal to limiting friction 10 N.

The free body diagram of two entities are shown in the figure.

a 1 = F K m 1 = 10 1 = 10 m / s 2 a 1 = F K m 1 = 10 1 = 10 m / s 2

a 2 = 50 F K m 2 = 50 10 2 = 20 m / s 2 a 2 = 50 F K m 2 = 50 10 2 = 20 m / s 2

The relative acceleration of 1 (block) with respect 2 (plank) is :

a 12 = a 1 a 2 a 12 = a 1 a 2

a 12 = 10 20 = 10 m s 2 a 12 = 10 20 = 10 m s 2

Thus, block has relative acceleration in the opposite direction to the direction of individual accelerations i.e. it is directed towards left. Now, initial velocity is zero and total displacement is 5m. Applying equation of motion for constant acceleration, we have :

x = u t + 1 2 a t 2 x = u t + 1 2 a t 2

5 = 0 + 1 2 X 10 t 2 5 = 0 + 1 2 X 10 t 2

t = 10 10 = 1 s t = 10 10 = 1 s

Problem 3 : In the arrangement shown, a force of 4mg, acts on the plank of mass “4m”. The coefficient of friction between block “A” and plank “B” is 0.25, whereas friction between plank and the surface underneath is negligible. Find the acceleration of “A” with respect to ground and its acceleration with respect to block “B”.

Solution : External force acts on the plank “B”. As such, it has tendency to move to right with respect to block “A”. The friction on the plank is, therefore, directed in the opposite direction i.e. towards left.

According to Newton’s third law, the friction acts on block in equal measure but in opposite direction i.e. towards right. We, however, need to know the nature of friction to proceed with the force analysis.

Here, limiting friction is :

F S = μ m g = 0.25 X 2 m g = m g 2 F S = μ m g = 0.25 X 2 m g = m g 2

We see here that friction is the only external force on block “A”. We assume here that friction is less than limiting friction (if assumption is wrong then we would get wrong result). In this condition, plank and block will together with a common acceleration given by :

a = F 6 m = 4 m g 6 m = 2 g 3 a = F 6 m = 4 m g 6 m = 2 g 3

The external force on the block would, then, be :

F A = 2 m a = 2 m X 2 g 3 = 4 m g 3 F A = 2 m a = 2 m X 2 g 3 = 4 m g 3

Clearly, external force on block “A” is greater than limiting friction, which is not possible as friction is the only external force on it. Friction, we know, can not exceed limiting friction. It means our assumption that block and plank move together has been wrong. It further means that two entities move with relative motion. As such, friction between them is kinetic friction (about equal to limiting friction).

Now, we know the magnitude and direction of friction force operating between the surface and hence, we can draw the free body diagrams of each of the entities as shown in the figure.

The acceleration of the block with respect to ground, “ a A a A ”, is given by :

a A = F S 2 m = m g 2 2m = g 4 a A = F S 2 m = m g 2 2m = g 4

The acceleration of plank with respect to ground, “ a B a B ”, is given by :

a B = 4 m g - F S 4 m = 4 m g m g 2 4 m = 7 g 8 a B = 4 m g - F S 4 m = 4 m g m g 2 4 m = 7 g 8

The relative acceleration of block “A” with respect to plank, “ a AB a AB ”, is :

a A B = a A a B = g 4 7 g 8 = 5 g 8 a A B = a A a B = g 4 7 g 8 = 5 g 8

The relative acceleration of “A” with respect to be is, thus, directed in the opposite direction of the motion.

## One external force each on block and plank

Problem 4 : A block of mass 4 kg is placed on a plank of mass 8 kg. At an instant, two external forces are applied on them as shown in the figure. The coefficient of friction for surfaces between block and plank is 0.5, whereas friction between plank and underneath surface is negligible. Find the friction between block and plank.

Solution : Considering block, the maximum static friction between block and plank is :

F S = μ N = μ m g = 0.5 X 4 X 10 = 20 N F S = μ N = μ m g = 0.5 X 4 X 10 = 20 N

The block and the plank will have relative motion with respect to each other when friction between the surfaces is kinetic friction (about equal to limiting friction). If friction between them is less than limiting static friction, then two entities will move together.

Here, we do not know the magnitude of friction. So we do not know whether two entities move together or not. Let us assume that they move together. If our assumption is wrong, then friction as determined from the force analysis under the assumed condition will equal to limiting friction, otherwise less than it.

Now, net external force on the combined body is towards left. Let the common acceleration be, “a”, and let the friction between block and plank be “ F F F F ”. The free body diagrams of block and plank are shown in the figure.

a = F F 5 4 = 40 F F 8 a = F F 5 4 = 40 F F 8

Solving for “ F F F F ”, we have :

8 F F 40 = 160 4 F F 8 F F 40 = 160 4 F F

12 F F = 200 12 F F = 200

F F = 16.66 N F F = 16.66 N

Thus, we see that friction at the interface is actually less than the limiting friction. Our assumption, therefore, was correct and hence, friction between the surface is 16.66 N.

## Block with initial velocity

Problem 5 : A plank “B” of mass “2m” is placed over smooth horizontal surface. A block “A” of mass “m” and having initial velocity “ v 0 v 0 ” is gently placed over the plank at one of its end as shown in the figure. If the coefficient of friction between “A” and “B” is 0.5, find the acceleration of “B” with respect to “A”.

Solution : We see here that block and plank has certain initial relative velocity. It means that friction between “A” and “B” is kinetic friction. The friction acts opposite to the velocity of the block “A”. On the other hand, friction is the only external force on the plank and acts opposite to the friction on “A”. Clearly, it is the friction that accelerates plank "B".

Let the magnitudes of accelerations of the block and the plank be “ a A a A ” and “ a B a B ” respectively with respect to ground. The free body diagrams, showing the forces on the block and the plank, are shown here.

The magnitudes of accelerations are :

a A = μ m g m = μ g a A = μ m g m = μ g

a B = μ m g 2 m = μ g 2 a B = μ m g 2 m = μ g 2

Two accelerations are in opposite direction. The relative acceleration of “B” with respect to “A” is :

a B A = a B a A = μ g 2 μ g = 3 μ g 2 a B A = a B a A = μ g 2 μ g = 3 μ g 2

The relative acceleration is in the same direction as that of the acceleration of block “B” i.e. towards right.

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