In this page, you can find the complete solutions of the third exercise of Composition and Resolution of Concurrent Forces chapter from Basic Mathematics Grade XI book published and distributed by Sukunda Pustak Bhawan.

In the above-mentioned book, Composition and Resolution of Concurrent Forces is the 21st chapter and has two exercises only. Out of which, this is the solution of the third exercise.


Check: Basic Mathematics Grade 11 (Sukunda Publication) Guide:

Grade 11 Basic Mathematics by Sukunda Pustak Vawan Notes and Solutions | Nepal


Disclaimer:

Answers mentioned here are not solved by teachers. These are the solutions written by a student of Grade 11. Answers are all correct. However, the language or process of solving the questions might be informal and in examinations, you might have to add little more language and show more calculations than what has been done here. So, we highly encourage you to view these solutions as guide rather than just copying everything mentioned here.

1) Three forces acting on a particle are in equilibrium; the angle between the first and second is 90$^o$ and that between the second and third is 120$^o$; find the ratio of the forces.

Solution:

We take the first force (A) along the X-axis. The second force (B) is 90$^o$ to the first force, i.e., along Y-axis. The third force (C) is 120$^o$ to the second force, which means it makes an angle 210$^o$ with positive X-axis.

X-component of the forces should result in 0 (equilibrium condition)

$\rm A \cos 0^O + B \cos 90^O + C \cos 210^O  = 0$

$\rm \Rightarrow A - \frac{\sqrt{3}}{2} C = 0$

$\rm \Rightarrow C = \frac{2}{\sqrt{3}} A$ - (i)

Y-component of the forces should result in 0 (equilibrium condition)

$\rm A \sin 0^O + B \sin 90^O + C \sin 210 ^O = 0$

$\rm \Rightarrow B - \frac{1}{2} C = 0$

$\rm \Rightarrow B = \frac{1}{2} C$

$\rm \Rightarrow C = 2 B$ - (ii)

From equation (i) and (ii)

$\rm C = \frac{2}{\sqrt{3}}A = 2 B$

$\rm \frac{C}{2} = \frac{A}{\sqrt{3}} = B$

$\rm \frac{A}{\sqrt{3}} = B = \frac{C}{2}$

$\rm \therefore A:B:C = \sqrt{3}:1:2$


2) The sides AB and AC of a triangle ABC are bisected in D and E; show that the resultant of forces represented by BE and DC is represented in magnitude and direction by 3/2 BC.



3) Find a point within a quadrilateral such that, if it be acted on by forces represented by the lines joining it to the angular points of the quadrilateral, it will be in equilibrium.


4) The sides BC and DA of a quadrilateral ABCD are bisected in F and H respectively; show that if two forces parallel and equal to AB and DC act on a particle, then the resultant is parallel to HF and equal to 2 HF.


5) A heavy chain has weights of 10 and 16 kg. attached to its ends and hangs in equilibrium over a smooth pulley. If the greatest tension of the chan in 20 kg.wt., find the weight of the chain.

Solution: