Physics · Laws Of Motion · NEET
Include only the forces acting ON your chosen body: its weight mg (always straight down), the normal reaction N from any surface (perpendicular to that surface), tension T (along the string, away from the body), friction f (along the surface, opposing relative sliding), and any applied push or pull. Leave OUT the forces this body exerts on other objects (by Newton's third law those act on the OTHER body, not this one). Also leave out the surface, string, or wall themselves — only their force arrows appear.
No. ma is NOT a force, so it never gets an arrow in the free body diagram. The FBD shows only real forces. After drawing it, you write Newton's second law as (sum of forces along a direction) = ma. The ma sits on the right side of the equation, not on the diagram. Drawing an ma arrow is a very common mistake that leads to double-counting.
No. The whole point of an FBD is to isolate the single body. You erase the surface, string and pulley and replace their effect with force arrows only: the surface becomes a normal reaction N (and friction f), the string becomes a tension T. This isolation is exactly NCERT's method: choose one system, then draw only that system and all forces on it from the surroundings.
Only two forces act: weight mg downward and normal reaction N upward. Since it is at rest, Newton's first law says the net force is zero, so N = mg. Many students wrongly add a third 'reaction to weight' arrow on the same block. That reaction (the pull of the block on the Earth) acts on the Earth, not on the block, so it does NOT appear in this FBD.
The normal reaction is always perpendicular to the contact surface and points AWAY from the surface, into the body. On a horizontal floor it points straight up. On an inclined plane it points perpendicular to the incline (not vertically up). Against a vertical wall it points horizontally away from the wall. Getting the normal direction right is essential before you resolve forces into components.
A block of mass 2 kg is placed on a rough inclined surface of coefficient of friction μ. It remains at rest. Taking g = 10 m/s², the net force on the block is:
A horizontal force of 10 N is applied to block A, which is in contact with and pushes block B on a frictionless horizontal surface. The masses of A and B are 2 kg and 3 kg respectively. The force exerted by block A on block B is:
A body of mass m is kept on a rough horizontal surface (coefficient of friction μ). A horizontal force is applied but the body does not move. The magnitude of the resultant F of the normal reaction and the frictional force satisfies:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is a picture where you draw only one body (as a dot or box) and show every force pushing or pulling on it as an arrow. You remove the surface, string and other bodies, keeping just the force arrows. It helps you write the correct equations to solve any mechanics problem.
Follow NCERT's method: (1) sketch the whole set-up, (2) choose one body as your system, (3) draw that body alone and mark all forces acting on it from the surroundings — weight, normal, tension, friction, applied force, (4) resolve forces along two convenient directions and apply F = ma. Do not include forces the body exerts on others.
Yes. Weight mg always acts vertically downward from the body's centre and is drawn in every FBD near the Earth, no matter the shape of the surface or the angle of an incline. It is the only force that never changes direction.
Because action and reaction always act on two DIFFERENT bodies. Your FBD is for one body only, so you draw just the force the surroundings exert on it. Its reaction acts on the other object and belongs in that object's FBD.
There is no fixed number — you include exactly the real forces touching or acting on that body. A block on a floor has 2 (weight, normal). On a rough incline with a string it may have 4 (weight, normal, friction, tension). Never add ma as a force.
