Physics · System Of Particles And Rotational Motion · NEET
Always take moments about the pivot (the wedge point). The wedge pushes up with a normal reaction N, but its distance from the pivot is zero, so N gives zero torque and drops out of the equation. This is why the pivot is the smartest point to choose: the unknown reaction force disappears and you are left with only the known weights. Taking moments about any other point would force you to also find N first.
For a uniform rod, its entire weight acts at its geometric centre (the midpoint). A 200 cm uniform rod has its weight acting at the 100 cm mark. This single downward force W = mg is placed at the centre of mass, and its moment arm is the distance from the pivot to that centre. Forgetting the rod's own weight is the most common mistake in wedge problems.
Stand the pivot in your mind. Any weight hanging on the LEFT of the pivot tries to rotate that side down — pick this as anticlockwise. Any weight on the RIGHT tries to rotate the right side down — clockwise. Weights on the same side of the pivot add together. Set (sum of anticlockwise moments) = (sum of clockwise moments) for balance.
N appears in the force balance (N = total downward weight) but NOT in the torque equation taken about the wedge, because its moment arm is zero there. For pure balancing questions you only need the torque equation, so you can ignore N completely and solve directly for the unknown mass.
No. The rod balances off-centre because extra masses are hung on it. The hung masses shift where the combined system balances. The rod's own weight still acts at its true middle (100 cm), but the added loads change the point where all clockwise and anticlockwise moments cancel, moving the balance point to the wedge position (40 cm).
A uniform rod (200 cm, 500 g) is balanced on a wedge at the 40 cm mark. A 2 kg mass hangs at the 20 cm mark and an unknown mass m hangs at the 160 cm mark. For equilibrium, m is: (g = 10 m/s²)
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The rod must be in rotational equilibrium: the sum of clockwise moments about the wedge equals the sum of anticlockwise moments. This is the principle of moments. The net force must also be zero, which fixes the normal reaction N equal to the total weight, but for balance questions only the moment condition is needed.
Every moment term is (mass × g × distance). Since g multiplies every single term on both sides of the balance equation, it divides out completely. That is why the final answer for an unknown mass does not depend on g, and you can even work in mass × distance units directly.
You must be consistent, but you do not have to use SI. Because both sides of the moment equation use the same units, you can keep grams and centimetres throughout as long as you use them everywhere. Converting to kg and m is safest for NEET to avoid slips.
A rod on a wedge involves only vertical forces (weights) and a single upward reaction, so one moment equation solves it. A rod leaning against a wall adds a horizontal wall reaction and floor friction, so you need all three equilibrium conditions: horizontal forces = 0, vertical forces = 0, and net torque = 0.
