Physics · System Of Particles And Rotational Motion · NEET
A smooth (frictionless) surface can only push along its own normal - the direction perpendicular to the surface. A vertical wall's normal points horizontally, so the wall force N_wall is horizontal. Friction (vertical along the wall) would need roughness, so a smooth wall gives zero vertical force. That is why only the floor can hold the ladder's weight up.
Use horizontal force balance. The only two horizontal forces are the wall normal N_wall (pushing away from wall) and the floor friction f (pushing toward the wall). Setting them equal gives f = N_wall. So first find N_wall from a torque equation, then friction equals it. In NEET 2025 this gave f = 100 root 3 N.
Take torque about the point where the most unknown forces act, usually the base on the floor. At the base, the floor normal and friction both pass through the pivot, so their torque is zero and they vanish from the equation. That leaves only the wall normal and the weight, giving one clean equation for N_wall.
For a uniform rod or ladder, the weight mg acts at the centre of mass, which is the middle of the length. Its torque arm about the base is half the horizontal span. Students often wrongly put mg at the contact point - always drop it at L/2 for a uniform ladder.
No. Angle with the wall and angle with the floor add up to 90 degrees. If the rod makes 60 degrees with the wall, it makes 30 degrees with the floor. Read the wording carefully, because sin and cos swap. In the NEET 2025 rod, 60 degrees with the wall means the horizontal reach uses cos 60 and vertical height uses sin 60 when measured from the wall side.
A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall, making 60 degrees with the wall; its lower end rests on a rough horizontal floor. The friction force exerted by the floor is: (g = 10 m/s^2)
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Net force must be zero (so it does not slide or fall) and net torque must be zero (so it does not rotate). In 2D this means three scalar equations: horizontal forces = 0, vertical forces = 0, and torques about any chosen point = 0.
Because the floor's normal force and friction both act at the bottom, taking torque there makes their moment arms zero. They disappear from the torque equation, leaving a single equation with just the wall force and the weight - the fastest route to the answer.
A rough wall adds a vertical friction force along the wall. Then vertical balance becomes N_floor + f_wall = mg, and the problem has more unknowns. NEET usually keeps the wall smooth so the wall force is purely horizontal, which is the standard case.
At the point of slipping, floor friction reaches its maximum f = mu x N_floor. Combining this with the torque equation gives a condition on the angle, typically tan(theta with floor) = 1/(2 mu) for a uniform ladder with a smooth wall. Below that angle the ladder slips.
For a uniform ladder with weight only at its centre, length cancels out of the torque equation, so the forces depend on the mass, the angle and g - not on the length. Length matters only when extra loads sit at specific marks along the ladder.
