NEET 2025 — Physics solution

NEET2025Physics

A uniform rod of mass $20$ kg and length $5$ m leans against a smooth vertical wall making an angle of $6 0^{\circ}$ with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is (take $g = 10$ m/s $^{2}$ ):

A
100 N
B
$1003$ N
C
200 N
D
$2003$ N

Solution

Vertical equilibrium: $N_{2} = m g = 200$ N. The rod makes $6 0^{\circ}$ with the wall, so it makes $3 0^{\circ}$ with the vertical and $6 0^{\circ}$ with the wall — equivalently, with the floor it makes $3 0^{\circ}$ . Wait, more carefully: angle with the wall is $6 0^{\circ}$ means angle with the floor is $3 0^{\circ}$ . The printed solution takes the angle in the torque equation as $6 0^{\circ}$ measured at the top, consistent with the figure.

Taking torques about the top end A (where the wall contacts the rod): the wall normal $N_{1}$ passes through A, so it contributes no torque.

$f_{s} L cos 6 0^{\circ} + m g \cdot \frac{L}{2} sin 6 0^{\circ} = N_{2} L sin 6 0^{\circ}$ .

$f_{s} \cdot \frac{1}{2} + 200 \cdot \frac{1}{2} \cdot \frac{3}{2} = 200 \cdot \frac{3}{2}$ .

$\frac{f _{s}}{2} = \frac{200 3}{2} - \frac{100 3}{2} = \frac{100 3}{2}$ , so $f_{s} = 1003$ N.

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