Equilibrium Of Concurrent Forces
Equilibrium of concurrent forces refers to the state where the vector sum of all forces acting on a body is zero, resulting in no net force and thus no acceleration. This fundamental principle in mechanics ensures that a body remains at rest or moves with constant velocity. It involves analyzing forces acting at a single point (point of concurrency) and applying vector addition principles to determine balance or imbalance. Understanding the equilibrium of concurrent forces is crucial for board exams and others like JEE Main and NEET. over the last ten years nine questions have been asked in JEE main and two questions have been asked in NEET.
Concurrent Forces
If all the forces working on a body are acting on the same point then they are said to be concurrent.
Three forces will be in equilibrium if they are represented by three sides of a triangle taken in order.
For equilibrium,
$\begin{aligned}
& \sum \overrightarrow{F_{n e t}}=0 \\
& \text { or } \sum \vec{F}_x=0, \sum \vec{F}_y=0, \sum \vec{F}_z=0
\end{aligned}$
Lami’s theorem
It states that for three concurrent forces in equilibrium,
$ \frac{F_1}{\sin \alpha}=\frac{F_2}{\sin \beta}=\frac{F_3}{\sin \gamma}$
For More Information On Equilibrium of Concurrent Forces, Watch The Below Video:
Solved Example Based On Equilibrium of Concurrent Forces
Now let’s understand the above law by some solved examples.
Example 1:Three forces start acting simultaneously on a particle moving with velocity $\vec{v}$ . These forces are represented in magnitude and direction by the three sides of a triangle ABC (as shown ). The article will now move with velocity
1) Less than $\bar{v}$
2) Greater than $\bar{v}$
3) $|\bar{v}|_{\text {in the direction of the largest force BC }}$
4) $\bar{v}$ remains unchanged
Solution:
Equilibrium of concurrent forces -
Three forces will be in equilibrium if they are represented by three sides of a triangle taken in order.
$\text { As Net force is zero. So } \bar{v} \text { remains unchanged. }$
Hence, the answer is option (4).
Example 2: A body of weight 100 N is suspended with the help of strings as shown in the figure. The tensions between T1 and T2 will be
1) 73.2 and 89.65
2)53.4 and 65.7
3)34.7 and 45.8
4)47.3 and 78.7
Solution:
Free body diagram of block
Resolution of forces along the horizontal direction gives:
$
\begin{aligned}
& T_1 \cos 30^{\circ}=T_2 \cos 45^{\circ} \\
& T_1 \times \frac{\sqrt{3}}{2}=T_2 \times \frac{1}{\sqrt{2}} \\
& \text { or, } T_2=\frac{\sqrt{3}}{\sqrt{2}} T_1
\end{aligned}
$
Resolution of forces along vertical direction gives:
$
\begin{aligned}
& T_1 \sin 30^{\circ}+T_2 \sin 45^{\circ}=100 \mathrm{~N} \\
& T_1 \times \frac{1}{2}+\sqrt{\frac{3}{2}} T_1 \times \frac{1}{\sqrt{2}}=100 \mathrm{~N} \\
& T_1=73.2 \mathrm{~N} \\
& T_2=89.65 \mathrm{~N}
\end{aligned}
$
Hence, the answer is option (1).
Example 3: If two concurrent forces A and B acting on a point are 200 N and 300 N. What is the magnitude of the resultant force (in Newtons), if it makes an angle of 50o with each force?
1) 455.12
2) 471.08
3) 400.56
4) 405.5
Solution:
Given:
$
\begin{aligned}
& \Theta=50^{\circ} \text {, force, }(A)=200 \mathrm{~N} \text {, force, }(B)=300 \mathrm{~N} \\
& \text { Resultant force, }\left(F_{\text {net }}\right)=\sqrt{A^2+B^2+2 A B \cos \Theta} \\
& F_{\text {net }}=\sqrt{A^2+B^2+2 A B \cos 50^{\circ}} \\
& \Rightarrow \sqrt{(200)^2+(300)^2+2 \times 200 \times 300 \times 0.62} \\
& =455.12 \mathrm{~N}
\end{aligned}
$
Hence, the answer is option (1).
Example 4: What makes forces concurrent?
1) Their lines of action must never cross each other
2) Their lines of action must all pass through the same point
3) They must act on at least two different objects
4) They must act in the same direction
Solution:
Concurrent Forces- If all the forces working on a body are acting on the same point then they are said to be concurrent.
A set of point forces is considered concurrent if all the lines of action of those forces pass through the same point.
Hence, the answer is option (2).
Example 5: A light spring balance hangs from the hook of the other light spring balance and a block of mass M Kg hangs from the former one. Then the true statement about the scale reading is :
1) both the scales read M Kg each
2)the scale of the lower one reads M kg and of the upper one zero
3)the reading of the two scales can be anything but the sum of the reading will be M kg
4)both the scales read M/2 kg
Solution:
Equilibrium of Concurrent Forces-If all the forces working on the body are acting on the same point then they are said to be concurrent.
where,
$\begin{aligned}
& \sum \overrightarrow{\mathrm{F}}_{\text {net }}=0 \\
& \sum \overrightarrow{\mathrm{F}}_{\mathrm{x}}=0, \sum \overrightarrow{\mathrm{F}}_y=0, \sum \overrightarrow{\mathrm{F}}_0=0
\end{aligned}$
hree forces will be in equilibrium if they are represented by three sides of triangle taken in order.
Hence, both the scales read M kg each
Hence, the answer is the option (1).
Summary
Concurrent forces are two or more forces working in parallel on an object simultaneously, but they may act in various directions of push or pull. For example, consider the following: picture a few people who pull or push a box from different sides. To determine what happens to the box, we add up all the forces acting on it. This combined force is called the resultant force. We find the resultant force by knowing both the magnitudes and direction of the forces being acted upon. We apply knowledge of concurrent forces to understand the motion of objects when several forces are acting and, importantly in the field of construction and sports, where usually things are being pushed, pulled, or kept in position by a variety of forces.
Frequently Asked Questions (FAQs)