Before starting this lesson, it may be helpful to study the basics of vectors.

In analyzing problems which involve forces acting on objects, it is very useful to make diagrams for each of the bodies involved, showing all the forces acting on it. Such diagrams are called Force diagrams or free body diagrams. Commonly seem forces are:

1) Gravitational Force. Usually exerted by the earth on objects.

2) Friction. Will always work to prevent slippage

3) Normal force between surfaces and objects. Newton’s third law: every action has an equal an opposite reaction.

4) Tension in a string.

Example 1) Free body diagram for a two block system at rest

Let us try to make a free body diagram for the mass block on top first. There are only two forces acting on it- gravity pulling it downwards and a normal force being applied by the other block. Let us call this normal force N_1

Normal forces can be quite confusing at first, but easy once the concept becomes clear. Normal force is the force which two bodies in contact exert on each other. In our given example, the block ‘m’ exerts the force N_1 on block ‘M’ (pushing it from the top)  and the block ‘M’ in return exerts an equal and opposite force on ‘m’, as seen in the free body diagram above.

Note that in a free body diagram, only the body in question is drawn, along with the forces acting on it. Surrounding bodies are not drawn to avoid confusion.

Let us look at the free body diagram for the block ‘M’ now

As discussed, the block ‘m’ exerts a normal force N_1 on the block ‘M’. The weight of the block also pulls it down. Both these forces are counterbalanced by the normal force exerted by the supporting surface N_2. Of course, the block ‘M’ also exerts an equal and opposite force on the supporting surface.

Since the bodies are at rest here, and there is no tendency to move, there is no friction in this case. Since the bodies are at rest, all forces are balanced out. This also follows from Newton’s second law- if acceleration is zero, net force must also be zero. Mathematically speaking:

F = ma= m0 =0

Mg+N_2=N_1

mg=N_2

Example 2) Mass on a string and pulley

A mass is being supported by a string which passes over a friction-less pulley. A Force F is applied to the other end of the string to keep the whole system at rest.

Here are the free body diagrams for the mass and the other end of the string to which the force is being applied.

T=mg

T=F

What would happen in case the Force at the other end of the string is applied at an angle instead of vertically downward? Since the pulley is friction-less, it makes no difference to the analysis. Only the force acting on the pulley changes.

Example 3) Wedge and mass system

A mass m is kept over a wedge of mass M and let go freely. All surfaces are friction-less. Both the mass and the wedge are movable. Analyze the forces acting on both bodies.

Intuitively speaking, the block will slip down along the wedge surface to the right, which will push the wedge to the left. There are no external forces present and friction is absent. This is a slightly complicated situation but can be analysed easily with the help of free body diagrams and Newton’s laws of motion.

In general, while attacking problems of this kind, it helps to follow the following steps-

Step 1) Draw the Free body Diagrams

Step 2) Choose a convenient x and y axes.

In our example, there are two options- either choose the horizontal-vertical x-y axis system, or choose the x axis to be along the wedge. The second option is not a good choice in this case because the wedge itself is moving (although it can be useful to analyze the motion of the mass relative to the wedge).

So, let’s choose the x and y axes as shown

Step 3) Project all forces onto the x and y axes. Be careful with + and – signs!

Forces acting on the wedge

Along x: -N_2\cdot \sin \theta

Along y: N_1-Mg-N_2\cos\theta

Forces acting on the mass m

Along x: N_2\sin \theta

Along y: N_2\sin \theta

We know that there is no acceleration along the y axis. So the net force must be also equal to zero. Using this we get

N_2=\frac{mg}{\cos\theta}

N_1=(M+m)g

Force acting on the wedge along x axis: -mg\tan\theta

Force acting on m along x axis: mg\tan \theta

Using these, we can easily derive the acceleration for both masses.

Understanding Static and Kinetic friction:

When there is a force trying to cause a movement but there is no actual movement, the force is called static friction. This force can vary between zero and \mu_sN. Once the driving force is enough to cause motion, the frictional force is \mu _kN\mu _k is usually less than \mu _s. We can graph the relationship between the driving force and the force of friction as below:

Example 4) In the pulley and mass system shown, mass m is just enough to keep the mass M from sliding down the wedge surface. The coefficient of static friction is \mu_s. Find m in terms of \mu _s, \theta and M. Note that the wedge itself is fixed.

In this problem its best to choose the x axis along the wedge since the motion of M is likely to be along that direction only (the wedge is fixed). Friction acts to prevent motion, that is up along the wedge (it is given the wedge is just prevented from sliding down).

Since mass m is stationary, the forces balance. Thus

T =mg

Also M does not move along the y axis or the x axis

N=Mg\cos\theta

mg + \mu _sMg\cos\theta=Mg\sin\theta

m=M(\sin\theta-\mu _s\cos\theta)

Let us look at another similar problem

Example 5) A block is moving on an inclined plane making an angle 45 degrees with the horizontal and the coefficient of friction is $latex\mu $. The force required to just push it up the inclined place is 3 times the force required to just prevent it from sliding down. If we define N = 10\mu. then N is (JEE 2011)

Again, we will chose the x axis to be along the wedge since motion is possible in this direction only. Let us call the force required to just push it up the place F_1 and the force required to just prevent it from sliding down as F_2. We need to draw the free body diagrams in both cases:

In the first case, friction acts to prevent F_1 from driving the mass up the plane. In the second case, friction helps F_2 to balance out gravitational force. In both cases

mg\cos\theta=N

Balancing the forces along the x axis for both cases (remember that there is no acceleration hence net force is zero)

F_1 = mg\sin\theta + \mu N = mg(\sin\theta+\mu\cos\theta)

F_2 = mg\sin\theta - \mu N = mg(\sin\theta-\mu\cos\theta)

Dividing the equations, we get

\frac{F_1}{F2}=\frac{(\sin\theta+\mu\cos\theta)}{(\sin\theta-\mu\cos\theta)}=\frac{(\tan\theta+\mu)}{(\tan\theta-\mu)}=3

Solve to get \mu =0.5 and N = 5. Keep in mind that this N has nothing to do with the normal force. This is simply a quantity defined in the question as 10 times \mu

Example 6) A block of mass m is on an inclined plane of angle \theta The coefficient of friction between the block and the plane is \mu and \tan\theta > \mu. The block is held stationary by applying a force P parallel to the plane. The direction of force pointing up the plane is taken to be positive. As P is varied from P_1 =mg(\sin\theta-\mu\cos\theta) to P_2 =mg(\sin\theta+\mu\cos\theta), the frictional force versus P graph will look like.

(JEE 2010)

This is very similar to the previous example.

Example 6) In the given figure, find tensions in each cord and the weight of the suspended body, if T is 10N.

Draw a free body diagram for the point where the three cords meet.

Since the arrangement is at rest, we can balance forces along the x and y axes.

10 \cos60=T_1\cos30

10\sin60+T_1\sin30 =W

Solve to get T_1= \frac{10}{\sqrt3}, W = \frac{20}{\sqrt3}

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