
Determine the point of application of force, when forces of \[20N\]and $30N$ are acting on a rod as shown in figure.

Answer
484.5k+ views
Hint: Force and moment analysis are required here for finding the point of application of force.
The net moment about any point due to all individual forces is equal to the moment of net force acting on the body is to be used here.
Use the point as the axis from which line of action of force is passing to automatically cancel its moment.
Complete step by step answer:
Moment is defined as product or force and perpendicular distance from the axis of rotation.
From the figure the net force acting on the body is
$\left( 30-20 \right)N$ =$10N$which is acting downwards
Let us choose the axis of rotation from where we will take the moments passing through C point.
We will be using the concept that moments due to individual forces about C and that of net force about C is equal.
Moments due to individual force =moment due to \[20N\]+ moment due to $30N$
Moment due to \[20N\]=$0$
Moment due to $30N$=$30N\times 20m$(it is distance from C point) ………… (1)
It is clockwise moment
Let the distance of net force from C be $x$
Moment due to net force=$10x$………. (2)
It is clockwise moment
Equating (1) and (2)
$\Rightarrow 10x=600$
$x=60m$ from point C.
So, the point of application of force has come out to be $x=60m$ from point C.
Note: Moment about an axis measures the tendency of the force to rotate the body along the given axis.
Choosing the axis is the most crucial part while taking moments.
Choose that axis from which maximum forces are passing so that their moment automatically becomes $0$.
The net moment about any point due to all individual forces is equal to the moment of net force acting on the body is to be used here.
Use the point as the axis from which line of action of force is passing to automatically cancel its moment.
Complete step by step answer:
Moment is defined as product or force and perpendicular distance from the axis of rotation.
From the figure the net force acting on the body is
$\left( 30-20 \right)N$ =$10N$which is acting downwards
Let us choose the axis of rotation from where we will take the moments passing through C point.
We will be using the concept that moments due to individual forces about C and that of net force about C is equal.
Moments due to individual force =moment due to \[20N\]+ moment due to $30N$
Moment due to \[20N\]=$0$
Moment due to $30N$=$30N\times 20m$(it is distance from C point) ………… (1)
It is clockwise moment
Let the distance of net force from C be $x$
Moment due to net force=$10x$………. (2)
It is clockwise moment
Equating (1) and (2)
$\Rightarrow 10x=600$
$x=60m$ from point C.
So, the point of application of force has come out to be $x=60m$ from point C.
Note: Moment about an axis measures the tendency of the force to rotate the body along the given axis.
Choosing the axis is the most crucial part while taking moments.
Choose that axis from which maximum forces are passing so that their moment automatically becomes $0$.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Trending doubts
Which one is a true fish A Jellyfish B Starfish C Dogfish class 11 biology CBSE

State and prove Bernoullis theorem class 11 physics CBSE

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

In which part of the body the blood is purified oxygenation class 11 biology CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells
