A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of 60° 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/s²)

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q22 – NEET-UG 2025

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Home Physics Laws of MotionQuestion #221

A uniform rod of mass

20

kg and length

5

m leans against a smooth vertical wall making an angle of

60^{\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}

)

Clue 1 of 3: Static Equilibrium

For a body to be in static equilibrium, both translational and rotational equilibrium conditions must be satisfied.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Identify Forces and Angles✦ Active

Draw a free-body diagram of the rod. The forces acting are: weight ( $m g$ ) acting at the center of mass, normal force from the floor ( $N_{f}$ ), friction force from the floor ( $f$ ), and normal force from the wall ( $N_{w}$ ). The rod makes an angle of $60^{\circ}$ with the vertical wall, which means it makes an angle of $90^{\circ} - 60^{\circ} = 30^{\circ}$ with the horizontal floor.

Step 2: Apply Translational Equilibrium Conditions○ Expand

For vertical equilibrium, the sum of vertical forces is zero:

N_{f} - m g = 0 ⟹ N_{f} = m g = 20 kg \times 10 {m/s}^{2} = 200 N

For horizontal equilibrium, the sum of horizontal forces is zero:

N_{w} - f = 0 ⟹ N_{w} = f

💡 Teacher's Secret Hint

The wall is smooth, so there is no friction from the wall. The friction force from the floor acts towards the wall to prevent slipping.

Step 3: Apply Rotational Equilibrium Condition○ Expand

Take torques about the point where the rod touches the floor (let's call it point A). The torques due to $N_{f}$ and $f$ about point A are zero. The forces causing torque are $m g$ and $N_{w}$ . The center of mass is at $L / 2$ from A.

Torque due to weight ( $m g$ ) (clockwise): $τ_{m g} = m g \times (L / 2) \cos 30^{\circ}$

Torque due to normal force from wall ( $N_{w}$ ) (counter-clockwise): $τ_{N_{w}} = N_{w} \times L \sin 30^{\circ}$

For rotational equilibrium, $\sum τ_{A} = 0$ :

N_{w} L \sin 30^{\circ} = m g (L / 2) \cos 30^{\circ}

Simplify the equation:

N_{w} \sin 30^{\circ} = (m g / 2) \cos 30^{\circ}

Solve for $N_{w}$ :

N_{w} = (m g / 2) \frac{\cos 30^{\circ}}{\sin 30^{\circ}} = (m g / 2) \cot 30^{\circ}

Substitute the values ( $m g = 200$ N, $\cot 30^{\circ} = \sqrt{3}$ ):

N_{w} = (200 N / 2) \times \sqrt{3} = 100 \sqrt{3} N

From Step 2, we know that the friction force $f = N_{w}$ . Therefore, the friction force exerted by the floor on the rod is $100 \sqrt{3}$ N.

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Physics Question 22 – NEET-UG 2025

Clue 1 of 3: Static Equilibrium

🧠 Full Solution Path