Physics · Laws Of Motion · NEET
The wedge is pushed forward, so the whole system accelerates. The block needs a sideways (horizontal) force to accelerate with the wedge. Only the normal force N from the smooth slope can supply this, because there is no friction. The horizontal part of N (N sinθ) pushes the block forward, and the vertical part (N cosθ) balances gravity. When a = g tanθ, N provides exactly the forward push needed AND holds the block up, so the block stays fixed on the slope.
Work in the ground frame. On the smooth block only two real forces act: weight mg (down) and normal N (perpendicular to slope). Horizontal: N sinθ = ma. Vertical: N cosθ = mg. Divide the first by the second: (N sinθ)/(N cosθ) = ma/mg, so tanθ = a/g. Therefore a = g tanθ. This is the single most tested wedge result for NEET.
Both give the same answer. Ground frame: use N sinθ = ma and N cosθ = mg (real forces only). Wedge frame (non-inertial): add a pseudo force ma pointing backward (opposite to the wedge's acceleration) on the block, then set the net force along the slope to zero: ma cosθ = mg sinθ, which again gives a = g tanθ. Pick whichever you find faster; the ground frame needs no pseudo force so it is safer for beginners.
In the wedge frame, the pseudo force on the block is ma and it points opposite to the wedge's acceleration — so if the wedge accelerates to the right, the pseudo force on the block points to the left (backward). Down the slope, gravity pulls it while the pseudo force pushes it up the slope; balancing these along the incline gives the stationary condition.
On a FIXED smooth incline the block always slides down with a = g sinθ, because nothing pushes it sideways. On an ACCELERATING wedge, the sideways push from the wedge (through N) can hold the block in place. If a is too small the block slides down; if a is too large it slides up; only at a = g tanθ does it stay still.
A block of mass m is placed on a smooth inclined wedge ABC of inclination θ. The wedge is given a horizontal acceleration 'a' towards the right. The relation between a and θ for the block to remain stationary on the wedge is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The wedge must be pushed horizontally with acceleration a = g tanθ, where θ is the angle of the smooth wedge. At this value the normal force alone holds the block up and drags it along, so it does not slide.
If a is smaller than g tanθ, the forward push is not enough, so the block slides DOWN the slope relative to the wedge. If a is larger than g tanθ, the block slides UP the slope.
No. The standard NEET problem uses a SMOOTH (frictionless) wedge, so only weight and normal force act. This is why the clean result a = g tanθ appears. Add friction only if the question clearly gives a coefficient μ.
From N cosθ = mg, we get N = mg/cosθ = mg secθ. Notice this is LARGER than the mg cosθ you get on a fixed incline, because the wedge is also pushing the block sideways.
It combines Newton's laws, free body diagrams and non-inertial frames (pseudo force) in one problem, and the exact a = g tanθ result has been asked directly (NEET 2018). It is a fast, high-return concept if you memorise the tan rule.
