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Question: A ladder of length \( L \) leans against a wall at an angle of \( \theta \) from the horizontal, as ...

A ladder of length LL leans against a wall at an angle of θ\theta from the horizontal, as shown in figure. The normal force FN{F_N} applied from the ground on the ladder applies what torque about the ladders center of mass?

\left( A \right){F_N}\dfrac{L}{2} \\\ \left( B \right){F_N}L\cos \theta \\\ \left( C \right){F_N}L\sin \theta \\\ \left( D \right){F_N}\dfrac{L}{2}\cos \theta \\\

Explanation

Solution

Hint : In order to solve this question, we are going to draw a figure showing all the forces that are present on the ladder at the different points in the vertical and the horizontal directions. Then, by defining the torque, we can calculate the torque on the ladder due to the forces acting on it.
The torque on the ladder is given as
τ=FH×r\tau = {F_H} \times r
Where FH{F_H} is the horizontal component of the force
And rr is the perpendicular distance from the center of mass of the rod.

Complete Step By Step Answer:
First of all, we need to start off by drawing a free body diagram for this configuration of the ladder, which gives the different forces with directions that are acting on this.

The direction of the normal force is shown in the figure above. The force FN{F_N} can be divided into two components, FNsin⁡θ{F_N}\sin \theta the vertical component is along the ladder, and FNcos⁡θ{F_N}\cos \theta the horizontal component is perpendicular to the ladder.
The torque about the ladder’s center of mass is given as:
τ=FH×r\tau = {F_H} \times r
Where FH{F_H} is the horizontal component of the force
And rr is the perpendicular distance from the center of mass of the rod.
\tau = {F_N}\cos \theta \times \dfrac{L}{2} \\\ \Rightarrow \tau = {F_N}\dfrac{L}{2}\cos \theta \\\
Hence, option (D)FNL2cos⁡θ\left( D \right){F_N}\dfrac{L}{2}\cos \theta is the correct answer.

Note :
The torque is defined as the moment of the force that makes a rod or ladder to rotate. Mathematically, the torque is the product of the force and the perpendicular distance from the center of mass of the rod. Although the torque is unable to rotate the ladder completely but it may cause a display.