Free Body Diagrams for AP Physics B and C

“To give pleasure to a single heart by a single act is better than a thousand heads bowing in prayer”

– Mahatma Gandhi

As AP Physics B and C aspirants you need to master the technique of drawing free body diagrams (FBD) without mistakes. You might have encountered many situations, especially in mechanics, where you were required to draw correct free body diagrams. Drawing free body diagrams at the level expected of you is ordinarily simple. Yet there may be many among you who have genuine doubts in this area.

A free body diagram of an object is one in which all the forces acting on the object are pictorially represented. The object may be the part of a complex system. But in the free body diagram of an object, you need to show the object alone (represented by a simple geometrical figure such as a dot or a small square or a small rectangle or the outline of the object) along with the forces represented by directed line segments (arrows). The length of the arrows, as usual in vector representation, should represent the magnitude of the force while the direction of the arrow should indicate the direction of the force. Since the free body diagram is a pictorial representation of the object along with the forces acting on it, used for simplifying your analysis of a problem, you need not worry about the exact length of the arrow and its exact direction. But the labeling of the magnitude and direction of the known forces should be correct.

In AP Physics free response questions often you may be asked to represent an object (such as a skier moving along a ramp or a body floating on the surface of a liquid) by a dot and to draw and label the forces that act on the object. Understand that you are being prompted to draw a simple free body diagram.

We will now draw the free body diagrams of some typical objects and systems in typical situations so that you will become more confident for facing your AP Physics Exam.

(1) A box of mass M at rest on a horizontal table:

The box can be represented by a small rectangle in the free body diagram as shown in FBD (i) in the figure. The forces acting on the box are the gravitational pull (weight) F_Gr and the normal force F_Normal exerted by the surface of the table_. These are of equal magnitude, but opposite in direction (F_Gr being downwards and F_Normal upwards. Surely, F_Gr = Mg.

[Even if you dispense with the symbol F_Gr for the gravitational pull and just write the weight Mg as label for it, your free body diagram will be alright].

If you were asked to represent the box by a dot, you would draw the free body diagram as shown in FBD (ii).

You know that the weight of the box acts through the centre of gravity of the box where as the normal force is exerted at the contact surface. You can draw the force vectors taking these into consideration; but there is no harm in showing them as in FBD (i) since the body has neither translational motion nor rotational motion since the equal and opposite forces have their lines of action coinciding.

When you are asked to represent the object by a dot, the question setter is in fact making things simpler for you.

(2) A block of mass M at rest on an incline:

In this case three forces act on the object. They are:

(i) The weight Mg of the block

(ii) The normal force F_Normal exerted by the incline and

(iii) The force of friction F_fr preventing the block from sliding down

The object is represented by a small rectangle in the free body diagram shown in FBD (i) in the figure. The frictional force F_fr acts along the surface of the block while the weight of the block acts through the centre of gravity of the block. The normal force in this case is opposite to the component of the weight normal to the inclined surface. Since the block does not rotate, the three forces acting on it have to be concurrent and so you need not be confused as to how to show the lines of action of the three forces. If you are asked to represent the block by a dot as shown in FBD (ii) you will be relieved of confusion.

[Note that the component of weight normal to the inclined surface is Mg cosθ where θ is the inclination of the plane (to the horizontal) and hence the magnitude of the normal force F_Normal is Mg cosθ. The frictional force F_fr is equal and opposite to the component (Mg sinθ) of the weight (of the block) parallel to the incline, as long as the block is at rest. If the block is just beginning to slide down, F_fr = μ_s Mg cosθ where μ_s is the coefficient of limiting friction (coefficient of static friction). If the block is in motion, F_fr = μ_k Mg cosθ where μ_k is the coefficient of kinetic friction. You need not show these details in the free body diagram. But you need to know these things in solving problems].

(3) A block of mass M sliding down an incline under gravity (No other external force applied):

The free body diagram will be the same as that shown in (2) above. But in this case the three forces are not balanced and so you have to show a smaller magnitude for the frictional force F_fr so that there is a net force acting down the plane to move the block

If you are asked take into account the viscous drag force F_D due to air, you have to show it along the direction of F_fr.

(4) A toy boat of mass m₁ carrying pebbles of mass m₂ floats on the surface of water in a jar:

In this case the object is the toy boat carrying glass beads. The total mass of the object is m₁ + m₂. Two forces act on the object. They are:

(i) The weight (m₁ + m₂)g of the object acting vertically downwards

(ii) The force of buoyancy F_Buoy acting vertically upwards.

You may represent the object by a small rectangle and indicate the weight and the force of buoyancy respectively by downward and upward arrows of equal length as indicated in FBD (i). Since the floating object is in equilibrium, the two vectors have to be equal and opposite and their lines of action must coincide.

[The weight acts through the centre of gravity of the object where as the force of buoyancy acts through the centre of buoyancy (centre of gravity of the displaced liquid). But you may ignore such details].

If you represent the object by a dot, things become quite simple and you get the free body diagram as shown in FBD (ii).

(5) An iron block of mass M suspended (from a rigid support) using a string and partially (or fully) submerged in a liquid.

In this case three forces act on the object (iron block). They are:

(i) The real weight Mg of the block acting vertically downwards

(ii) The force of buoyancy F_Buoy acting vertically upwards

(iii) The tension T in the string, acting vertically upwards

The free body diagram of the iron block can be drawn either as in FBD (i) or as in FBD (ii).

Since the iron block is in equilibrium, you have to draw the magnitudes of the forces such that magnitude of Mg = Magnitude of (F_Buoy + T).

[Free body diagrams you draw in your answer sheets are rough sketches and hence no examiner will expect you to show the exact magnitudes of the forces].

(6) A skier of mass M skiing down a frictionless incline with a drag force F_d due to air opposing her motion

Three forces act on the object (skier). They are:

(i) The weight Mg of the skier acting vertically downwards

(ii) The drag force F_D acting parallel to and up the incline

(ii) The normal force F_Normal exerted by the incline.

Usually you will be asked to represent the skier by a dot. You will draw the free body diagram of the skier as shown.

[If there is friction along the incline, you will have to show the frictional force also in same direction as that of the drag force].

(7) A cricket ball of mass m moving through air after being hit by a batsman (with negligible air resistance)

Since air resistance is negligible, the only force acting on the ball is its weight mg acting vertically downwards. The free body diagram is therefore very simple as shown in the adjoining figure.

[The direction of motion of the ball is immaterial in this simple case of negligible air resistance and you get the same free body diagram at all positions of the ball along its trajectory in air. But, if air resistance (drag force due to air) F_D is significant and is to be considered, you will have to show it directed opposite to the direction of motion of the ball. This direction will change continuously with time if the ball is following a parabolic path in air. If you are asked to draw the free body diagram at the instant the ball is at the highest point of its parabolic trajectory, the drag force will be horizontal and opposite to the direction of motion of the ball. But if the ball is projected vertically up, it will be momentarily at rest at the highest point and hence the viscous drag force will be zero there].

(8) Blocks A and B of masses M₁ and M₂ (M₁ > M₂) tied to the ends of a string (of negligible mass) passing over a frictionless fixed pulley, with block A resting on the ground and block B hanging in air

You are required to draw the free body diagrams of block A as well as block B.

The arrangement of the blocks and pulley is shown in the figure. The required free body diagrams are also shown.

Three forces act on block A. They are:

(i) The weight m₁g of the block acting vertically downwards

(ii) The normal force F_Normal exerted by the ground, acting vertically upwards

(iii) The tension T in the string, acting vertically upwards

The important thing to be noted in the case of block A is that the normal force F_Normal exerted on it by the ground is not of magnitude M₁g but is less than M₁g. The value is (M₁– M₂) g and the tension T in the string is M₂ g

Two forces act on block B. They are:

(i) The weight M₁g of the block acting vertically downwards

(ii) The tension T in the string, acting vertically upwards

[Since the same segment of string connects the two masses, the tension in the string has the same value].

The above situation can be suitably modified in your question paper. For instance, the block A can be a man of mass 70 kg standing on the ground and holding the end of the string and the block B can be a boy of mass 30 kg suspended from the other end of the string!

[If both blocks A and B were in air, the system of masses will move with an acceleration since there is a net driving force (M₁–M₂)g acting on the system. The free body diagram of block A will be similar to that of block B and the normal force F_Normal will be absent. The magnitude of the tension T in the string will be greater than M₁g, but less than M₁g].

(9) A tethered hydrogen balloon of mass M kept stationary in air. The mass of the rope is negligible.

You are required to draw the free body diagram of the balloon.

Three forces act on the balloon. They are:

(i) The weight Mg of the balloon, acting vertically downwards.

(ii) The force of buoyancy F_Buoy acting vertically upwards.

(iii) The tension T in the string, acting vertically downwards.

Since the mean density of the hydrogen balloon is less than that of air, the force of buoyancy will be greater than the weight of the balloon. The weight Mg of the balloon and the tension T in the rope together prevent the balloon from rising up.

The free body diagram of the balloon can be drawn as shown in FBD (i). If you were asked to represent the balloon by a dot, you would draw the free body diagram as shown in FBD (ii).

Newton’s Laws of Motion for AP Physics- Equations to be Remembered

We shall require a substantially new manner of thinking if mankind is to survive.

– Albert Einstein

Even though most of you will be remembering the important points in connection with Newton’s Laws of Motion, it will be better to have a glance at the following:

(1) Inertia is a basic property of any material body, by virtue of which it resists any change in its state of rest or of uniform motion.

(2) A force is required to change the state of rest or of uniform motion of a body. The resultant force acting on a body at rest or in uniform motion is zero. [Note that a body in uniform motion has uniform velocity].

(3) Newton’s second law is mathematically expressed as

F_net = ma where F_net is the net (resultant) force and a is the acceleration.

This can be written in terms of momentum p as

F_net = dp/dt

Often we write this as F= dp/dt, understanding that F is indeed the net force.

[Remember that the mass m can be treated as constant only at speeds negligible compared to the speed of light].

(4) Impulse given to an object (by a force) = F∆t where F is the force and ∆t is the time for which the force acts.

If the force is not constant and it acts from the instant t₁ to the instant t₂, we have

Impulse = _t1 ∫^t2 Fdt

This gives the area under the force-time graph between the ordinates corresponding to the times t₁ and t₂ (Shaded area in fig.)

Since force F = ∆p/∆t where ∆p is the change in momentum during the time ∆t, we can write

Impulse = F∆t = (∆p/∆t) ∆t = ∆p

Thus impulse = change of momentum

(5) Motion in a lift

The weight of a body of mass ‘m’ in a lift can be remembered as m(g-a) in all situations if you apply the proper sign to the acceleration ‘a’ of the lift. The acceleration due to gravity ‘g’ always acts vertically downwards and its sign may be taken as positive. The following cases can arise in this context:

(i) Lift moving down with acceleration of magnitude ‘a’:

In this case ‘a’ also is positive and the weight is m(g-a) which is less than the real weight of the body (when it is at rest).

(ii) Lift moving up with acceleration:

In this case ‘a’ is negative and the weight is m[g-(-a)] = m(g+a).

(iii) Lift moving down with retardation (going to stop while moving down):

In this case also ‘a’ is negative and the weight is m[g-(-a)] =m(g+a) which is greater than the actual weight.

(iv) Lift moving up with retardation (going to stop while moving up):

In this case ‘a’ is positive and the weight is m(g-a)

(v) Lift moving up or down with uniform velocity:

In this case ‘a’ is zero and the weight is mg.

(vi) Lift moving down with acceleration of magnitude ‘g’ (falling freely under gravity as is the case when the rope carrying the lift breaks):

In this case ‘a’ is positive and the weight is m(g-g) which is zero.

If you have a clear idea of the weight of a body in a lift, you will be able to use it in other similar situations as well (for instance, the motion of bodies connected by a string passing over a pulley).

[We will discuss conservation of momentum separately in due course].

(6) Friction

The force of friction, F_fric ≤ μN where μ is the coefficient of friction and N is the normal reaction (normal force).

In the adjoining figure, a body being pulled along a horizontal surface by a horizontal force F is shown. The frictional force F_fric is maximum when the body just begins to move and is called limiting force of static friction (F_s)_max so that we have

(F_s)_max = μ_s N. This gives the value of the coefficient of static friction μ_s as

μ_s = (F_s)_max /N

When the body slides along the surface, the friction called into play is called kinetic friction. The force of kinetic friction F_k is less than the above limiting value (F_s)_max and the corresponding coefficient of kinetic friction μ_k is less than μ_s. We have

μ_k = F_k/N

If the body rolls along the surface, The friction called into play is called rolling friction which is much less than kinetic friction.

Angle of friction λ is the angle between the the normal force N and the resultant reaction S. As shown in the figure, the resultant reaction is the resultant of the normal force N and the frictional force F_fric. Since tan λ = F_fric/N, it follows that

μ = tan λ

A body of mass m placed on a ramp (inclined plane) is shown in the adjoining figure. The component mg sinθ of the weight mg of the body is the force trying to move the body down the plane. The normal reaction is the reaction (force) opposing the component mg cosθ of the component of the weight of the body normal to the inclined plane. The frictional force F_fric is opposite to the component mg sinθ (of the weight of the body) parallel to the plane. Note that friction is a self adjusting force up to its maximum value (F_s)_max and if the body shown in the figure is at rest, F_fric is just sufficient to balance the component mg sinθ (of the weight of the body).

If the inclination of the plane is gradually increased from a small value, the body placed on it will begin to slide down when the angle is equal to the angle of friction, λ. The angle of repose is therefore equal to the angle of friction.

Often you may be asked to draw a free body diagram (FBD), indicating the forces acting on the body. In the case of the body placed on the inclined plane, the free body diagram is as shown. The body is represented by a dot. The forces to be shown are the weight mg of the body, the normal force N (equal to mg cosθ) exerted by the inclined surface on the body and the frictional force F_fric since they are the actual forces acting on the body. Don’t worry about the components mg sinθ and mg cosθ of the weight. The real force is the weight mg which we have shown already. We consider the components just for the convenience of explanation. The normal reaction (force) offered by the surface and the frictional force between the body and the surface are to be accommodated in addition to the weight of the body.

[Note that if you want, you can draw the FBD showing the components mg sinθ and mg cosθ of the weight of the body. But in that case you will not show the weight mg in the FBD].

If the inclined plane is smooth, the frictional force F_fric will be absent in the free body diagram.

If the body is held on a smooth incline by a spring fixed to the incline, the spring force Kx has to be shown in place of the frictional force F_fric. Here K is the spring constant and x is the elongation (or compression as the case may be) of the spring.

If the body moves down the incline and the viscous drag force (air resistance) is significant, that too is to be shown up the incline.

In the next post we will discuss questions in this section. Meanwhile, find some useful multiple choice questions (with solution) here.