Composition and Resolution of Concurrent Force Exercise 21.3 | Basic Mathematics Solution [NEB UPDATED]

Exercise 21.3

1) Three forces acting on a particle are in equilibrium, the angle between the first and second is 90° and that between the second and third is 120° find the ratio of the forces

Solution:

Let P,Q and R be the forces acting at O. The angle between third and first = 360° - 90° = 150°.

Using sine law,

Or, $\frac{{\rm{P}}}{{{\rm{sin}}120\infty }}$ = $\frac{{\rm{Q}}}{{{\rm{sin}}150\infty }}$ = $\frac{{\rm{R}}}{{{\rm{sin}}90\infty }}$

Or, $\frac{{\rm{P}}}{{\frac{{\sqrt 3 }}{2}}}$ = $\frac{{\rm{Q}}}{{\frac{1}{2}}}$ = $\frac{{\rm{R}}}{1}$.

Or, $\frac{{\rm{P}}}{{\sqrt 3 }}$ = $\frac{{\rm{Q}}}{1}$ = $\frac{{\rm{R}}}{2}$

So, P:Q:R = $\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 $\frac{3}{2}\overrightarrow {{\rm{BC}}} $.

Solution:

Using vector addition,

Or,${\rm{\: }}\overrightarrow {{\rm{BE}}} $ = $\overrightarrow {{\rm{BC}}}  + \overrightarrow {{\rm{CE}}} $

= $\overrightarrow {{\rm{BC}}}  + \frac{1}{2}\overrightarrow {{\rm{CA}}} $ ….(i)

Or, $\overrightarrow {{\rm{DC}}} $ = $\overrightarrow {{\rm{DB}}}  + \overrightarrow {{\rm{BC}}} $

= $\frac{1}{2}\overrightarrow {{\rm{AB}}}  + \overrightarrow {{\rm{BC}}} $ ….(ii)

Adding (i) and (ii)

Or, $\overrightarrow {{\rm{BE}}}  + \overrightarrow {{\rm{DC}}} $ = 2$\overrightarrow {{\rm{BC}}} $ + $\frac{1}{2}\overrightarrow {{\rm{CA}}} $ + $\frac{1}{2}\overrightarrow {{\rm{AB}}} $ = 2$\overrightarrow {{\rm{BC}}} $ + $\frac{1}{2}$($\overrightarrow {{\rm{CA}}} $ + $\overrightarrow {{\rm{AB}}} $).

= 2$\overrightarrow {{\rm{BC}}} $ + $\frac{1}{2}\overrightarrow {{\rm{CB}}} $ = 2$\overrightarrow {{\rm{BC}}} $ + $\frac{1}{2}$($\overrightarrow {{\rm{CA}}} $ + $\overrightarrow {{\rm{AB}}} $).

= 2$\overrightarrow {{\rm{BC}}} $ + $\frac{1}{2}\overrightarrow {{\rm{CB}}} $ = 2$\overrightarrow {{\rm{BC}}} $ – $\frac{1}{2}\overrightarrow {{\rm{BC}}} $ = $\frac{3}{2}\overrightarrow {{\rm{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.

Solution:

Let O be a point within the quadrilateral ABCD such that,

Or, $\overrightarrow {{\rm{OA}}} $ + $\overrightarrow {{\rm{OB}}} $ + $\overrightarrow {{\rm{OC}}} $ + $\overrightarrow {{\rm{OD}}} $ = 0

Or $\overrightarrow {{\rm{OA}}} $ + $\overrightarrow {{\rm{OB}}} $ = 2$\overrightarrow {{\rm{OM}}} $ ….(i)

Where M is the middle point of AB,

Again, $\overrightarrow {{\rm{OC}}} $ + $\overrightarrow {{\rm{OD}}} $ = 2$\overrightarrow {{\rm{ON}}} $ …(ii)

Where N is the middle point of CD.

From (i) amd (ii)

Or, 2$\overrightarrow {{\rm{OM}}} $ + 2$\overrightarrow {{\rm{ON}}} $ = $\overrightarrow {{\rm{OA}}} $ + $\overrightarrow {{\rm{OB}}} $ + $\overrightarrow {{\rm{OC}}} $ + $\overrightarrow {{\rm{OD}}} $

Or, 2$\overrightarrow {{\rm{OM}}} $ + 2$\overrightarrow {{\rm{ON}}} $ = 0

So, $\overrightarrow {{\rm{OM}}}  + \overrightarrow {{\rm{ON}}} $ = 0

Ie. $\overrightarrow {{\rm{OM}}} $ and $\overrightarrow {{\rm{ON}}} $ are equal and opposite.

So, O is the middle point of MN.

Similarly, we can show that O is the middle point of PQ.

So, O is the point of intersection of the line joining the middle points of the opposite sides of a quadrilateral.

 

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.

Solution:

Or, $\overrightarrow {{\rm{AB}}} $ = $\overrightarrow {{\rm{AF}}}  + {\rm{\: }}\overrightarrow {{\rm{FB}}} $.

= $\overrightarrow {{\rm{AH}}}  + \overrightarrow {{\rm{HF}}}  + \overrightarrow {{\rm{FB}}} $ ….(I)

Or, $\overrightarrow {{\rm{DC}}} $ = $\overrightarrow {{\rm{DF}}}  + \overrightarrow {{\rm{FC}}} $.

= $\overrightarrow {{\rm{DH}}} $+ $\overrightarrow {{\rm{HF}}} $ + $\overrightarrow {{\rm{FC}}} $. ….(ii)

Adding (i) and (ii),

Or, $\overrightarrow {{\rm{AB}}} $ + $\overrightarrow {{\rm{DC}}} $ = $\left( {\overrightarrow {{\rm{AH}}}  + \overrightarrow {{\rm{DH}}} } \right)$ + 2$\overrightarrow {{\rm{HF}}} $ – ($\overrightarrow {{\rm{FB}}} $ + $\overrightarrow {{\rm{FC}}} $) = 0 + 2.${\rm{\: }}\overrightarrow {{\rm{HF}}} $ + 0

= 2$\overrightarrow {{\rm{HF}}} $ (So, $\overrightarrow {{\rm{AH}}} $ and $\overrightarrow {{\rm{DH}}} $ are equal and opposite etc.).

 

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 chain is 20 kg wt., find the weight of the chain.

Solution:

Let A and B ne the ends of the chain to which the weight of 10kgs, and 16kgs are attached and C be the highest point of the chain on the pulley. As,20kgs is the greatest tension of the chain,

So,

Weight of AC + 10kgs = 20kgx

And Weight of BC + 16kgs = 20kgs

Adding we get,

Weight of (AC + BC) + 26 = 40.

Or, Weight of chain = 40 – 26 = 14kgs.

 

6) Two men carry a weight 50N between two strings fixed to the weight one string is inclined at 30° to the vertical and the other at 60", find the tension of each string.

Solution:

Let OA and OB be two strings inclined at an angle 30° and 60° with the vertical CO respectively and T₁ and T₂ be the tensions along OA and OB respectively. The weight 50N is vertically downwards from O.

Now, by Lami’s theorem, we have,

Or, $\frac{{{{\rm{T}}_1}}}{{\sin \left( {{\rm{BOC}}} \right)}}$ = $\frac{{{{\rm{T}}_2}}}{{\sin \left( {{\rm{COA}}} \right)}}$ = $\frac{{50}}{{\sin \left( {90\infty } \right)}}$.

Or, $\frac{{{{\rm{T}}_1}}}{{{\rm{sin}}120\infty }}$ = $\frac{{{{\rm{T}}_2}}}{{{\rm{sin}}150\infty }}$ = $\frac{{50}}{1}$.

Or, $\frac{{{{\rm{T}}_1}}}{{\frac{{\sqrt 3 }}{2}}}$ = $\frac{{{{\rm{T}}_2}}}{{\frac{1}{2}}}$ = 50,

So, T₁ = 50 * $\frac{{\sqrt 3 }}{2}$ = 25$\sqrt 3 $. N.

And T₂ = 50 * $\frac{1}{2}$ = 25N.

 

7) A body weighing 4 N is supported by a string attached to a fixed point and is pulled from the vertical by a horizontal force of 3 N. Find the angle the string will make with the vertical and the tension of the string.

Solution:

Let $\angle $AOB = α. Let T be the tension of the sting,

Using Lami’s theorem,

Or, $\frac{{\rm{T}}}{{{\rm{sin}}90\infty }}$ = $\frac{3}{{\sin \left( {180\infty  - \alpha } \right)}}$ = $\frac{4}{{\sin \left( {90\infty  + \alpha } \right)}}$.

Or, $\frac{{\rm{T}}}{1}$ = $\frac{3}{{{\rm{sin}}\alpha }}$ = $\frac{4}{{{\rm{cos}}\alpha }}$

So, tanα= $\frac{3}{4}$.

i.e. sinα = $\frac{3}{5}$

or, α= sin^-1$\left( {\frac{3}{5}} \right)$.

T = $\frac{3}{{{\rm{sin}}\alpha }}$ = $\frac{3}{{\frac{3}{5}}}$ = 5N.

 

8) A body of weight 65 N is suspended by two strings of lengths 5 and 12 m attached to two points in the same horizontal line whose distance apart is 13m; find the tensions of the strings.

Solution:

OA = 5m, OB = 12m, AB = 13m

OA² + OB² = 25 + 144.

= 169 = (13)²

= AB²

So, $\angle $AOB = 90°.

Let $\angle $OBA = θ,

Then, $\angle $BOC = 90° - θ.

And $\angle $AOC = θ

Let T₁and T₂ be the tensions of the strings OA and OB respectively,

Using Lami’s theorem.

Or, $\frac{{{{\rm{T}}_1}}}{{\sin \left( {180\infty  - 90\infty  + \theta } \right)}}$ = $\frac{{{{\rm{T}}_2}}}{{\sin \left( {180\infty  - \theta } \right)}}$ = $\frac{{65}}{{{\rm{sin}}90\infty }}$

Or, $\frac{{{{\rm{T}}_1}}}{{{\rm{cos}}\theta }}$ = $\frac{{{{\rm{T}}_2}}}{{{\rm{sin}}\theta }}$ = $\frac{{65}}{1}$.

So, T₁ = 65.cosθ = 65 * $\frac{{12}}{{13}}$ = 60N.

T₂ = 65sinθ = 65 * $\frac{5}{{13}}$ = 25N.

 

9) The ends of an inelastic and weightless string 0.17 m long are attached to two points 0.13 m apart in the same horizontal line and a weight of 4 N is attached to string 0.05 m from one end. Find the tension in each portion of the string.

Solution:

OA = 0.05m, OB = 0.12m

AB = 0.13m

OA² + OB² = (5cm)² + (12cm)²

= 25 + 144.

= 169 = (13cm)² = AB²

So, $\angle $ AOB = 90°.

If $\angle $OBC = θ, then $\angle $BOC = 90° - θ.

And $\angle $AOC = θ.

Let T₁ and T₂ be the tensions of the strings OA and OB.

Using Lami’s theorem,

Or, $\frac{{{{\rm{T}}_1}}}{{\sin \left( {180\infty  - 90\infty  + \theta } \right)}} = \frac{{{{\rm{T}}_2}}}{{\sin \left( {180\infty  - \theta } \right)}} = \frac{4}{{{\rm{sin}}90\infty }}$.

Or, $\frac{{{{\rm{T}}_1}}}{{{\rm{cos}}\theta }} = \frac{{{{\rm{T}}_2}}}{{{\rm{sin}}\theta }} = \frac{4}{1}$.

So, T₁ = 4cosθ = 4 * $\frac{{12}}{{13}}$ = $\frac{{48}}{{13}}$N = 3.69N

T₂ = 4sinθ = 4 * $\frac{5}{{13}}$ = $\frac{{20}}{{13}}$N = 1.54N.

 

10) A uniform sphere of weight 3 N rests in contact with a smooth vertical wall. It is supported by a string whose length equals the radius of the sphere, joining a point on the surface of the sphere to a point of the wall. Calculate the tension in the string and the reaction of the wall.

Solution:

A is the point of contact of the sphere and the vertical wall. String BC = the radius OA = OB and $\angle $OAC = 90°.

So, OAB is an equilateral triangle,

So, $\angle $AOB = 60°.   [OB = BC and $\angle $OAC = 90°]

Let R be the normal reaction. The weight is vertical. Let T be the tension on the string,

Or, $\frac{{{{\rm{T}}_1}}}{{{\rm{sin}}90\infty }} = \frac{{\rm{R}}}{{{\rm{sin}}150\infty }} = \frac{3}{{{\rm{sin}}120\infty }}$

Or, $\frac{{{{\rm{T}}_1}}}{1} = \frac{{\rm{R}}}{{\frac{1}{2}}} = \frac{3}{{\frac{{\sqrt 3 }}{2}}}$.

So, T₁ = 2$\sqrt 3 $N = 2 * 1.73N = 3.46N.

R = $\sqrt 3 $N = 1.73N.

 

11) Forces P, Q, R acting along OA, OB, OC where O is the circumcenter of the triangle ABC, are in equilibrium, show that

Solution:

Or, $\angle $BOC = 2A, $\angle $COA = 2B, and $\angle $AOB = 2C.   [O is the centre of circum circle ABC]

Using Lami’s theorem, 

(i) $\frac{{\rm{P}}}{{{\rm{sin}}\angle {\rm{BOC}}}}$ = $\frac{{\rm{Q}}}{{{\rm{sin}}\angle {\rm{COA}}}}$ = $\frac{{\rm{R}}}{{{\rm{sin}}\angle {\rm{AOB}}}}$.

Or, $\frac{{\rm{P}}}{{{\rm{sin}}2{\rm{A}}}} = \frac{{\rm{Q}}}{{{\rm{sin}}2{\rm{B}}}} = \frac{{\rm{R}}}{{{\rm{sin}}2{\rm{C}}}}$.

Or, $\frac{{\rm{P}}}{{2{\rm{sinA}}.{\rm{cosA}}}}$ = $\frac{{\rm{Q}}}{{2{\rm{sinB}}.{\rm{cosB}}}}$ = $\frac{{\rm{R}}}{{2{\rm{sinC}}.{\rm{cosB}}}}$

Or, $\frac{{\rm{P}}}{{{\rm{acosA}}}} = \frac{{\rm{Q}}}{{{\rm{bcosB}}}} = \frac{{\rm{R}}}{{{\rm{ccosC}}}}$$\left[ {\frac{{\rm{a}}}{{{\rm{sinA}}}} = \frac{{\rm{b}}}{{{\rm{sinB}}}} = \frac{{\rm{c}}}{{{\rm{sinC}}}} = 2{\rm{R}}} \right]$

 

(ii) $\frac{{\rm{P}}}{{{\rm{acosA}}}} = \frac{{\rm{Q}}}{{{\rm{bcosB}}}} = \frac{{\rm{R}}}{{{\rm{ccosC}}}}$.

Or, $\frac{{\rm{P}}}{{\frac{{{\rm{a}}\left( {{{\rm{b}}^2} + {{\rm{c}}^2} - {{\rm{a}}^2}} \right)}}{{2{\rm{bc}}}}}} = \frac{{\rm{Q}}}{{\frac{{{\rm{b}}\left( {{{\rm{c}}^2} + {{\rm{a}}^2} - {{\rm{b}}^2}} \right)}}{{2{\rm{ca}}}}}} = \frac{{\rm{R}}}{{\frac{{{\rm{c}}\left( {{{\rm{a}}^2} + {{\rm{b}}^2} - {{\rm{c}}^2}} \right)}}{{2{\rm{ab}}}}}}$

Or, $\frac{{\rm{P}}}{{{{\rm{a}}^2}\left( {{{\rm{b}}^2} + {{\rm{c}}^2} - {{\rm{a}}^2}} \right)}} = \frac{{\rm{Q}}}{{{{\rm{b}}^2}\left( {{{\rm{c}}^2} + {{\rm{a}}^2} - {{\rm{b}}^2}} \right)}} = \frac{{\rm{R}}}{{{{\rm{c}}^2}\left( {{{\rm{a}}^2} + {{\rm{b}}^2} - {{\rm{c}}^2}} \right)}}$.

 

12) O is the orthocentre of the triangle ABC. Forces P, Q, R acting along OA, OB, OC are in equilibrium. Prove that: $\frac{{\rm{P}}}{{{\rm{BC}}}} = \frac{{\rm{Q}}}{{{\rm{CA}}}} = \frac{{\rm{R}}}{{{\rm{AB}}}}$.

Solution:

Or, $\angle $CBF = 90°- C.

Or, $\angle $BCF = 90° - B.

From triangle BOC,

So, 90° - C + 90° - B + $\angle $BOC = 180°.

So, $\angle $BOC = B + C = 180° - A.

Similarly, $\angle $COA = 180° - B.

And $\angle $AOD = 180° - C.

Using lami’s theorem,

Or, $\frac{{\rm{P}}}{{\sin \left( {180\infty  - {\rm{A}}} \right)}} = \frac{{\rm{Q}}}{{\sin \left( {180\infty  - {\rm{B}}} \right)}} = \frac{{\rm{R}}}{{\sin \left( {180\infty  - {\rm{C}}} \right)}}$.

Or, $\frac{{\rm{P}}}{{{\rm{sinA}}}} = \frac{{\rm{Q}}}{{{\rm{sinB}}}} = \frac{{\rm{R}}}{{{\rm{sinC}}}}$.

Or, $\frac{{\rm{P}}}{{\rm{a}}} = \frac{{\rm{Q}}}{{\rm{b}}} = \frac{{\rm{R}}}{{\rm{c}}}$.

Or, $\frac{{\rm{P}}}{{{\rm{BC}}}} = \frac{{\rm{Q}}}{{{\rm{CA}}}} = \frac{{\rm{R}}}{{{\rm{AB}}}}$.

 

13) 13 ABC is a triangle and D, E, F are the middle points of the sides BC, CA, AB respectively. Show that the forces represented by the straight lines AD, BE, CF acting at a point are in equilibrium.

Solution:

Or, $\overrightarrow {{\rm{AD}}} $ = $\overrightarrow {{\rm{AB}}}  + \overrightarrow {{\rm{BD}}} $ = $\overrightarrow {{\rm{AB}}} $ + $\frac{1}{2}\overrightarrow {{\rm{BC}}} $.

Or, $\overrightarrow {{\rm{BE}}} $ = $\overrightarrow {{\rm{BC}}}  + \overrightarrow {{\rm{CE}}} $ = $\overrightarrow {{\rm{BC}}} $ + $\frac{1}{2}\overrightarrow {{\rm{CA}}} $.

Or, $\overrightarrow {{\rm{CF}}} $ = $\overrightarrow {{\rm{CA}}} $ + $\overrightarrow {{\rm{AF}}} $ = $\overrightarrow {{\rm{CA}}} $ + $\frac{1}{2}\overrightarrow {{\rm{AB}}} $.

Adding,

Or, $\overrightarrow {{\rm{AD}}} $ + $\overrightarrow {{\rm{BE}}} $ + $\overrightarrow {{\rm{CF}}} $ = $\overrightarrow {{\rm{AB}}} $ + $\frac{1}{2}\overrightarrow {{\rm{BC}}} $ + $\overrightarrow {{\rm{BC}}} $ + $\frac{1}{2}\overrightarrow {{\rm{CA}}} $ + $\overrightarrow {{\rm{CA}}} $ + $\frac{1}{2}\overrightarrow {{\rm{AB}}} $.

= $\frac{1}{2}\left( {\overrightarrow {{\rm{AB}}}  + \overrightarrow {{\rm{BC}}}  + \overrightarrow {{\rm{CA}}} } \right)$ = $\frac{3}{2}\left( {\overrightarrow {{\rm{AC}}}  + \overrightarrow {{\rm{CA}}} } \right)$ = 0.

So, the forces represented by AD,BE and CF are in equilibrium.

 

14) A body of weight 20 N, which hangs by a string is pushed to one side by a horizontal force so that the string makes an angle of 60° with the vertical, find the horizontal force and the tension of the string.

Solution:

Or, $\angle $AOB = 60°.

Let P be the horizontal force and T be the tension of the string,

Using Lami’s theorem,

Or, $\frac{{\rm{P}}}{{{\rm{sin}}120\infty }}$ = $\frac{{\rm{T}}}{{{\rm{sin}}90\infty }}$ = $\frac{{20}}{{{\rm{sin}}150\infty }}$.

Or, $\frac{{\rm{P}}}{{\frac{{\sqrt 3 }}{2}}}$ = $\frac{{\rm{T}}}{1}$ = $\frac{{20}}{{\frac{1}{2}}}$.

Or, $\frac{{2{\rm{P}}}}{{\sqrt 3 }}$ = $\frac{{\rm{T}}}{1}$ = 40.

So, P = 20$\sqrt 3 $N, T = 40N.

 

15) A heavy chain of length 9m and weighing 18 N has a weight of 6 N attached to one end is in equilibrium hanging over a smooth peg. What length of the chain is on each side?

Solution:

Total length of the chain = 9m, AB = xm.

Length of the part of the chain AC = (9 – x)m

Total wt. of the chain = 18N.

Wt. of chain of length 1m = $\frac{{18}}{9}$ = 2N.

Wt. of chain of length xm = 2x.

Wt.of chain of length (9 – x)m = 2(9 – x).

By the question,

Or, 6 + 2x = 2(9 – x).

Or, 6 + 2x = 18 – 2x.

Or, 4x = 12.

So, x = 3m.

Length AB = xm = 3m, Length AC = (9 – x) = 6m.

 

16) A uniform plane lamina in the form of a rhombus, one of whose angles is 120°, is supported by two forces applied at the centre in the directions of the diagonals so that one side of the rhombus is horizontal, show that if P and Q be the forces and P be the greater. then P² = 3Q².

Solution:

ABCD is a rhombus with diagonals AC and BD intersecting O.

Or, $\angle $ABC = 120°, $\angle $BAD = 60°.

Let W be the weight of the rhombus. Along the diagonals AC and BD forces are applied so that AB is horizontal. WMN is vertical,

Or, $\angle $OBA = 60°.

So, $\angle $BOM = vert. opp $\angle $DON = 30°.

Similarly, $\angle $AOM = vert, opp $\angle $CON = 60°.

So, $\angle $DON <$\angle $CON, the greater force P must be along BD and Q along AC.

Or, $\frac{{\rm{P}}}{{\sin \left( {90\infty  + 30\infty } \right)}}$ = $\frac{{\rm{Q}}}{{\sin \left( {90\infty  + 60\infty } \right)}}$.

Or, sin150°.P = Q.sin120°.

Or, $\frac{1}{2}$.P = Q.$\frac{{\sqrt 3 }}{2}$.

So, P² = 3Q²