Answer
Verified
410.7k+ views
Hint: The ladder is in equilibrium against a wall, this means it is at rest or in uniform motion and no external forces act on it. We can apply Newton’s second law for forces acting in different directions and make equations. We can also apply equilibrium in its rotational motion and make equations for torque acting on it and solve the equations to calculate the value of forces.
Formulas used:
$\tau =F\times r$
${{N}_{1}}-F=0$
${{N}_{2}}-mg=0$
Complete answer:
The system of ladder is an isolated system, analyzing all the forces acting on it and resolving the forces into its components as follows
Since the ladder is on rest against the wall, the resultant of all forces acting on it will be 0.
Forces acting on the ladder in the x-direction
$\begin{align}
& {{N}_{1}}-F=0 \\
& \Rightarrow {{N}_{1}}=F \\
\end{align}$
Forces acting in the y-direction
$\begin{align}
& {{N}_{2}}-mg=0 \\
& \Rightarrow {{N}_{2}}=mg \\
\end{align}$
Now taking torque about point O, we have,
Since the ladder is in equilibrium, the net torque acting on it is 0.
Therefore, the torque is calculated as-
$\tau =F\times r$ - (2)
Here, $\tau $ is the torque
$F$ is the force acting on the ladder
$r$ is the distance from the axis
The axis is passing through point O
Distance of point of action of ${{N}_{1}}$ from the axis is calculated as-
From the triangle given in the above figure
$\begin{align}
& \dfrac{x}{L}=\sin \theta \\
& \Rightarrow x=L\sin \theta \\
\end{align}$
Distance of point of action of ${{N}_{2}}$ from the axis is calculated as-
$\begin{align}
& \dfrac{y}{L}=\cos \theta \\
& \Rightarrow y=L\cos \theta \\
\end{align}$
Distance of point of action of $mg$ from the axis is $\dfrac{L}{2}\cos \theta $
Therefore, from eq (2), the torque acting on the ladder is-
$\begin{align}
& {{N}_{1}}L\sin \theta +{{N}_{2}}L\cos \theta -mg\dfrac{L}{2}\cos \theta =0 \\
& \Rightarrow FL\sin \theta +mgL\cos \theta -mg\dfrac{L}{2}\cos \theta =0 \\
& \Rightarrow FL\sin \theta +mg\dfrac{L}{2}\cos \theta =0 \\
& \Rightarrow FL\sin \theta =-mg\dfrac{L}{2}\cos \theta \\
& \Rightarrow F=-\dfrac{mg}{2}\cot \theta \\
\end{align}$
Substituting values for ladder in the above equation, we get,
$\therefore F=-\dfrac{Mg}{2}\cot \theta $
Therefore,
(a). The normal reaction of the wall on the ladder is $\dfrac{Mg}{2}\cot \theta $.
(b). The normal reaction of the ground on the ladder is $Mg$.
(c). the force applied by the ground on the ladder is $\dfrac{Mg}{2}\cot \theta $
Note:
The negative sign indicates that the force is opposite to the assumed direction. The normal reaction is perpendicular to the surfaces and by Newton’s third law, it acts equal and opposite on both the surfaces. The frictional force resists the motion of a body; it is applied by rough surfaces. When an object is in equilibrium, no net force or torque acts on it.
Formulas used:
$\tau =F\times r$
${{N}_{1}}-F=0$
${{N}_{2}}-mg=0$
Complete answer:
The system of ladder is an isolated system, analyzing all the forces acting on it and resolving the forces into its components as follows
Since the ladder is on rest against the wall, the resultant of all forces acting on it will be 0.
Forces acting on the ladder in the x-direction
$\begin{align}
& {{N}_{1}}-F=0 \\
& \Rightarrow {{N}_{1}}=F \\
\end{align}$
Forces acting in the y-direction
$\begin{align}
& {{N}_{2}}-mg=0 \\
& \Rightarrow {{N}_{2}}=mg \\
\end{align}$
Now taking torque about point O, we have,
Since the ladder is in equilibrium, the net torque acting on it is 0.
Therefore, the torque is calculated as-
$\tau =F\times r$ - (2)
Here, $\tau $ is the torque
$F$ is the force acting on the ladder
$r$ is the distance from the axis
The axis is passing through point O
Distance of point of action of ${{N}_{1}}$ from the axis is calculated as-
From the triangle given in the above figure
$\begin{align}
& \dfrac{x}{L}=\sin \theta \\
& \Rightarrow x=L\sin \theta \\
\end{align}$
Distance of point of action of ${{N}_{2}}$ from the axis is calculated as-
$\begin{align}
& \dfrac{y}{L}=\cos \theta \\
& \Rightarrow y=L\cos \theta \\
\end{align}$
Distance of point of action of $mg$ from the axis is $\dfrac{L}{2}\cos \theta $
Therefore, from eq (2), the torque acting on the ladder is-
$\begin{align}
& {{N}_{1}}L\sin \theta +{{N}_{2}}L\cos \theta -mg\dfrac{L}{2}\cos \theta =0 \\
& \Rightarrow FL\sin \theta +mgL\cos \theta -mg\dfrac{L}{2}\cos \theta =0 \\
& \Rightarrow FL\sin \theta +mg\dfrac{L}{2}\cos \theta =0 \\
& \Rightarrow FL\sin \theta =-mg\dfrac{L}{2}\cos \theta \\
& \Rightarrow F=-\dfrac{mg}{2}\cot \theta \\
\end{align}$
Substituting values for ladder in the above equation, we get,
$\therefore F=-\dfrac{Mg}{2}\cot \theta $
Therefore,
(a). The normal reaction of the wall on the ladder is $\dfrac{Mg}{2}\cot \theta $.
(b). The normal reaction of the ground on the ladder is $Mg$.
(c). the force applied by the ground on the ladder is $\dfrac{Mg}{2}\cot \theta $
Note:
The negative sign indicates that the force is opposite to the assumed direction. The normal reaction is perpendicular to the surfaces and by Newton’s third law, it acts equal and opposite on both the surfaces. The frictional force resists the motion of a body; it is applied by rough surfaces. When an object is in equilibrium, no net force or torque acts on it.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Difference Between Plant Cell and Animal Cell
Give 10 examples for herbs , shrubs , climbers , creepers
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE