Physics · Laws Of Motion · NEET
No. N = mg is true ONLY for a body resting on a flat horizontal surface with no vertical push or pull. On an incline of angle θ, N = mg cos θ (smaller than mg). In a lift moving up with acceleration a, N = m(g + a). If you press down on the object, N increases; if you pull up, N decreases. Always find N from the equation of forces perpendicular to the surface, never assume N = mg.
The word 'normal' in physics means perpendicular. The contact force between two surfaces has two parts: the part perpendicular to the surfaces is called normal reaction (N), and the part parallel to the surfaces is called friction (f). By definition N is the perpendicular part, so it always points straight out of the surface, away from it.
No. Weight (mg) is the pull of gravity, acting downward toward the Earth's centre. Normal reaction is a contact push from the surface, acting perpendicular to that surface. They happen to be equal and opposite only on a flat horizontal surface at rest. On an incline or in an accelerating lift they are different in size and direction.
No, this is a common trap. Both weight and normal reaction act on the SAME body (the block), so they cannot be a Newton's-third-law pair. Action-reaction pairs act on DIFFERENT bodies. The true reaction to the surface's normal force N (surface on block) is the force the block presses on the surface (block on surface). The true reaction to the block's weight is the block's gravitational pull on the Earth.
On an inclined plane the normal force points perpendicular to the slope surface, tilted away from the vertical by the angle of the incline. It is NOT straight up. Because it is perpendicular to the slope, its magnitude is N = mg cos θ, where θ is the angle of the incline. Only the component of gravity perpendicular to the slope is balanced by N.
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:
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:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Yes. When contact is lost, N becomes zero. For example, at the top of a vertical circle at minimum speed, or when a lift falls freely (a = g downward), N = m(g − g) = 0. This is the state of apparent weightlessness.
If you push down at angle θ with force P, N = mg + P sin θ. If you pull up at angle θ, N = mg − P sin θ. In a lift accelerating up, N = m(g + a); accelerating down, N = m(g − a).
For a body sliding on a fixed surface, no. Because N is perpendicular to the surface and the motion is along the surface, the angle between N and displacement is 90 degrees, so work = N·s·cos 90 = 0.
Friction depends directly on N. Limiting (maximum) static friction is μsN and kinetic friction is μkN. If you get N wrong (for example using mg on an incline instead of mg cos θ), every friction value that follows will also be wrong.
