In this page, you can find the complete solutions of the second 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 three exercises only. Out of which, this is the solution of the second 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. A force equal to 10 N is inclined at an angle of 30 to the horizontal; find its resolved parts in a horizontal and vertical directions.

Solution:

Let $F = 10 N$

$\theta = 30^o$

We know,

Horizontal component $F_x= F cos \theta$

$= 10 * cos 30$

$= 10 \times \dfrac{\sqrt{3}}{2}$

$= 5 \sqrt{3} N$

Vertical component $F_y = F sin \theta$

$= 10 * sin 30 $

$= 10 * \dfrac{1}{2}$

$= 5 N$

2. Resolve a force equal to 50 N into two forces making angles of 60 and 45 with it in opposite sides.

Solution:

Let the two resolved forces be A and B and their angles with the original force be denoted by $\alpha$ and $\beta$, respectively.

Then, we have,

$A = \dfrac{50 \sin \alpha}{sin( \alpha + \beta)}$

$= \dfrac{50 \sin 60^o}{\sin (60 + 45)}$

$= \dfrac{50 \times \frac{ \sqrt{3}}{2} }{ \sin 105^o}$

$= \dfrac{25 \sqrt{3}}{ \frac{ \sqrt{3} + 1}{2 \sqrt{2}}}$

$= \dfrac{25 \sqrt{3} \times 2 \sqrt{2}}{\sqrt{3} + 1}$

$= \dfrac{50 \sqrt{6} }{ \sqrt{3} + 1} \times \dfrac{ \sqrt{3} - 1}{ \sqrt{3} -1 }$

$= \dfrac{50 \sqrt{6} ( \sqrt{3} - 1)}{3 - 1}$

$= 25 (3 \sqrt{2} - \sqrt{6})$ N

And

$A = \dfrac{50 \sin \alpha}{sin( \alpha + \beta)}$

$= \dfrac{50 \sin 45^o}{\sin (60 + 45)}$

$= \dfrac{50 \times \frac{ 1}{\sqrt{2}} }{ \sin 105^o}$

$= \dfrac{25 \sqrt{2}}{ \frac{ \sqrt{3} + 1}{2 \sqrt{2}}}$

$= \dfrac{25 \sqrt{2} \times 2 \sqrt{2}}{\sqrt{3} + 1}$

$= \dfrac{50 \times 2}{\sqrt{3} + 1} \times \dfrac{\sqrt{3} - 1}{\sqrt{3} - 1}$

$= \dfrac{100 (\sqrt{3} -1)}{3 -1}$

$= 50(\sqrt{3}-1)$ N

Hence, the required forces are $25 (3 \sqrt{2} - \sqrt{6})$ N and $50 ( \sqrt{3} -1 )$ N.


3. If a force P be resolved into two forces making angles of 45 and 15 with its direction; show that the latter force is $\dfrac{\sqrt{6}}{3} P$.

Solution:

Let the two resolved forces be respresented by A and B and their angle with the force P be $\alpha$ and $\beta$.

Then,

$A = \dfrac{P \sin \alpha}{sin(\alpha + \beta)}$

$A = \dfrac{P \sin 45^o}{sin(45+15)^o}$

$= \dfrac{P \times \dfrac{1}{\sqrt{2}}}{sin 60^o}$

$= \dfrac{P \times \dfrac{1}{\sqrt{2}}}{\dfrac{\sqrt{3}}{2}}$

$= \dfrac{P \times 2}{\sqrt{6}}$

$= \dfrac{P \times 2 }{\sqrt{6}} \times \dfrac{\sqrt{6}}{\sqrt{6}}$

$= \dfrac{P \times 2 \sqrt{6}}{6}$

$= \dfrac{\sqrt{6}}{3}P$

However, the latter force is B that makes angle $\beta$ with the given force. Just to make the answer matching with the book, we solved it accordingly.


About the Textbook:

Name: Basic Mathematics Grade XI
Author(s): D.R. Bajracharya | R.M. Shresththa | M.B. Singh | Y.R. Sthapit | B.C. Bajracharya
Publisher: Sukunda Pustak Bhawan (Bhotahity, Kathmandu)
Telephone: 5320379, 5353537
Price: 695 /- (2078 BS)