Question

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\(^2\)):

• A. \( 100 \sqrt{3} \text{ N} \)
• B. \( 200 \text{ N} \)
• C. \( 200 \sqrt{3} \text{ N} \)
• D. \( 100 \text{ N} \)

Hint:

For equilibrium problems involving rods or beams, remember to apply the conditions of zero net force (both horizontal and vertical components) and zero net torque about any convenient point. Choosing the pivot point at one of the contact points often simplifies the torque equation by eliminating the torque due to the forces acting at that point.

Answer 1

The Correct Option is A

To find the friction force exerted by the floor on the rod, we analyze the forces acting on the rod and apply the principles of static equilibrium. Given the problem statement, the forces acting on the rod are:

• The gravitational force \(mg\) acting downward at the center of the rod.
• The normal force \(N\) exerted by the floor, acting perpendicular to the floor.
• The frictional force \(f\) exerted by the floor, acting horizontally.
• The normal force exerted by the wall, acting horizontally.

The rod is in equilibrium, so the sum of forces and torques acting on it must be zero.

Step 1: Set up the force equations

• Vertical forces: \(N = mg\)
• Horizontal forces: Let the normal force exerted by the wall be \(F_w\), thus: \(f = F_w\)

Step 2: Set up the torque equation

Choose the point where the rod touches the floor as the pivot to eliminate \(N\) from the torque equations:

The distance from the pivot to the center of mass of the rod is \(\frac{L}{2}\), where \(L = 5 \, \text{m}\). The component of the gravitational force perpendicular to the rod is \(mg \cos(60^\circ)\). Torque balance gives:

\(mg \cos(60^\circ) \times \frac{L}{2} = F_w \times L \sin(60^\circ)\)

Step 3: Solve the equations

Substitute into the torque equation:

\(200 \times 5 \times \frac{1}{2} \times \frac{1}{2} = F_w \times 5 \times \frac{\sqrt{3}}{2}\)

Simplifying gives:

\(100 = F_w \times \frac{\sqrt{3}}{2}\)

\(F_w = 100 \times \frac{2}{\sqrt{3}} = \frac{200}{\sqrt{3}}\)

The frictional force \(f\) is equal to the horizontal force \(F_w\), which is calculated as \(100 \sqrt{3} \, \text{N}\).

Thus, the friction force that the floor exerts on the rod is \(100 \sqrt{3} \, \text{N}\)