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)

To find the friction force exerted by the horizontal floor on the rod, we can analyze the forces acting on the uniform rod in equilibrium.

Let us identify the given values:
Mass of the uniform rod, $m=20\text{kg}$
Length of the rod, $L=5\text{m}$
Angle made by the rod with the vertical wall, $\theta ={60}^{\circ }$
Acceleration due to gravity, $g=10{\text{m/s}}^2$

Let the vertical wall be along the y-axis (smooth) and the horizontal floor be along the x-axis (rough).
The forces acting on the rod are:
1. The weight of the rod, $W=mg=20\times 10=200\text{N}$, acting vertically downwards at its center of gravity (middle of the rod, at a distance of $L/2$ from either end).
2. The normal reaction force from the smooth vertical wall, $N_1$, acting horizontally (perpendicular to the wall).
3. The normal reaction force from the rough horizontal floor, $N_2$, acting vertically upwards.
4. The frictional force, $f$, exerted by the floor on the rod, acting horizontally towards the wall to prevent the rod from slipping outwards.

For the rod to be in translational equilibrium:
The net horizontal force must be zero:
$N_1-f=0\Rightarrow f=N_1$
The net vertical force must be zero:
$N_2-W=0\Rightarrow N_2=W=200\text{N}$

For the rod to be in rotational equilibrium:
The net torque about any point must be zero. Let us calculate the torque about the point of contact with the floor (let's call it point A):
The torque due to the normal force $N_2$ and frictional force $f$ about point A is zero since their lines of action pass through A.
The torque due to the weight $W$ (acting downwards at the midpoint, distance $L/2$) tends to rotate the rod clockwise:
${\tau }_W=W\times \left(\frac{L}{2}\sin \left(\theta \right)\right)$
The torque due to the normal reaction of the wall $N_1$ (acting horizontally at the top end, distance $L$) tends to rotate the rod counter-clockwise:
${\tau }_{N_1}=N_1\times \left(L\cos \left(\theta \right)\right)$

Equating the magnitudes of these opposing torques for rotational equilibrium:
$N_{1}L\cos \left(\theta \right)=W\frac{L}{2}\sin \left(\theta \right)$
We can cancel the length of the rod $L$ from both sides:
$N_1\cos \left(\theta \right)=\frac{W}{2}\sin \left(\theta \right)$
Solving for $N_1$:
$N_1=\frac{W}{2}\tan \left(\theta \right)$

Substitute the values $W=200\text{N}$ and $\theta ={60}^{\circ }$:
$N_1=\frac{200}{2}\tan \left({60}^{\circ }\right)$
$N_1=100\sqrt{3}\text{N}$

Since the friction force $f$ equals $N_1$:
$f=100\sqrt{3}\text{N}$

Thus, the friction force that the floor exerts on the rod is 100 √3N.