Composition and Resolution of Concurrent Force Exercise 21.1 | Basic Mathematics Solution [NEB UPDATED]

Exercise 21.1

1)In the following problems P N and Q N denote two component forces acting at an angle a and RN denotes their resultant

i) IF P=24, Q=7, α= 90, find R.

Solution:

R² = P² + Q² + 2Pcosα

= 24² + 7² + 2.24.7cos90° = 576 + 49 + 0 = 625.

So, R = 25N.

ii) IF P 13, R=14, α=90 find Q

Solution:

R² = P² + Q² + 2PQ cosα.

Or, 14² = 13² + Q² + 2.13.Qcos90°

Or, 196 = 169 + Q² + 0

So, Q = 3$\sqrt 3 $N.

iii) If Q=3, R=7, α=60 find P.

Solution:

R² = P² + Q² + 2.PQcosα

Or, 7² = 3² + Q² + 2.3.Q.cos60°.

Or, 49 = 9 + Q² + 3Q

Or, Q² + 3Q – 40 = 0

Or, (Q + 8)(Q – 5) = 0

So, Q = -8,5

So, Q ≠ -8.

So, Q = 5N.

iv) IF P=3, Q=5, R=7 find α.

Solution:

R² = P² + Q² + 2PQcosα.

Or, 7² = 3² + 5² + 2.3.5 cosα

Or, 49 – 9 – 25 = 30cosα.

Or, 15 = 30cosα.

Or, cosα = $\frac{1}{2}$.

So, α= 60°

2) Find the resultant of the two forces P=4 N and Q=3N

i) When P acts towards east and Q towards north

Solution:

α = 90°(north)

So, R = $\sqrt {{{\rm{P}}^2} + {{\rm{Q}}^2} + 2{\rm{PQcos}}\alpha } $ = $\sqrt {{4^2} + {3^2} + 0} $ = 5N.

And tanθ = $\frac{{\rm{Q}}}{{\rm{P}}}$ = $\frac{3}{4}$ [α = 90°]

So, θ = tan^-1$\left( {\frac{3}{4}} \right)$ with 4N.

ii) When P acts towards cast and Q towards north-west

Solution:

α = 135° (north-west)

So, R = $\sqrt {{{\rm{P}}^2} + {{\rm{Q}}^2} + 2{\rm{PQcos}}\alpha } $ = $\sqrt {{4^2} + {3^2} - 2.4.3.{\rm{\: }}\frac{1}{{\sqrt 2 }}} $

So, R = $\sqrt {26 - 12\sqrt2 } $,N

And θ = tan^-1$\left( {\frac{{{\rm{Qsin}}\alpha }}{{{\rm{P}} + {\rm{Qcos}}\alpha }}} \right)$ = tan^-1$\left( {\frac{{3.{\rm{\: }}\frac{1}{{\sqrt 2 }}}}{{4 - 3.{\rm{\: }}\frac{1}{{\sqrt 2 }}}}} \right)$

So, θ = tan^-1$\left( {\frac{3}{{4\sqrt 2  - 3}}} \right)$ with 4N.

 

3) Two forces P and 2P acting at a point, have the resultant √3P Find the angle between the two given forces.

Solution:

Let α be the angle between the forces,

So, ${\left( {\sqrt 3 {\rm{P}}} \right)^2}$ = P² + (2P)² + 2P.2P.cosα.

Or, 3P² = 5P² + 4P²cosα.

Or, 4P²cosα = -2P².

Or, cosα= $ - \frac{1}{2}$.

So, α = 120°.

 

4) Two forces whose magnitudes are P and $\sqrt 2 $P N act on a particle in directions inclined at an angle of 135^o to each other, find the magnitude and direction of the resultant.

Solution:

P = P, Q = P$\sqrt 2 $, α = 135°, R = ?.

R² = P² + Q² + 2PQcosα = P² + 2p² + 2P.$\sqrt 2 $P.cos135°.

So, R = P.

Tanθ = $\frac{{{\rm{P}}\sqrt 2 .{\rm{sin}}135\infty }}{{{\rm{P}} + {\rm{P}}\sqrt 2 {\rm{cos}}135\infty }}$ = $\frac{{{\rm{P}}\sqrt 2 \left( {\frac{1}{{\sqrt 2 }}} \right)}}{{{\rm{P}} + {\rm{P}}\sqrt 2 \left( { - \frac{1}{{\sqrt 2 }}} \right)}}$ = ∞.

So, θ= 90°.

 

5) Two forces acting at an angle of 45^o have a resultant equal to $\sqrt 10 $ N. if one of the force be √2 N, find the other force.

Solution:

α = 45°, R = $\sqrt {10} $N, P = $\sqrt 2 $N, Q = ?

Or, R² = P² + Q² + 2PQcosα.

Or, 10 = 2 + Q² + 2.$\sqrt 2 $.Q cos45°

Or, 10 = 2 + Q² + 2$\sqrt 2 $Q.$\frac{1}{{\sqrt 2 }}$.

Or, Q² + 2Q – 8 = 0

Or, (Q + 4)(Q – 2) = 0

SO, Q = -4,2

So, Q ≠ -4.

So, Q = 2N.

 

6) At what angle do forces equal to (P+Q) newtons and (P-Q) newtons act so that the resultant may be $\sqrt {{P^2} + {Q^2}} $?

Solution:

${\left( {\sqrt {{{\rm{P}}^2} + {{\rm{Q}}^2}} } \right)^2}$ = (P + Q)² + (P – Q)² + 2(P + Q)(P – Q)cosα

Or, P² + Q² = P² + 2PQ + Q² + P² – 2PQ + Q² + 2(P² – Q²).cosα

Or,  - P² – Q² = 2(P² – Q²)cosα.

So, cosα = $ - \frac{{{{\rm{P}}^2} + {{\rm{Q}}^2}}}{{2\left( {{{\rm{P}}^2} - {{\rm{Q}}^2}} \right)}}$

 

7) Find the magnitude of two forces such that if they act at right angles, their resultant is $\sqrt 10 $  N. whilst if they act an angle of 60", their resultant is √13 N

Solution:

P = ?, Q = ?, R = $\sqrt {10} $, α= 90°.

R² = P² + Q² + 2PQcos90°.

Or, 10 = P² + Q² + 0

So, P² + Q² = 10 …(i)

P = ?, Q = ?, R = $\sqrt {13} $, α= 60°.

R² = P² + Q² + 2PQ.cos60°.

Or, 13 = P² + Q² + 2PQ.$\frac{1}{2}$.

Or, 13 = 10 + PQ  [From (i)]

So, PQ= 3.

Or, (P + Q)² = P² + Q² + 2PQ = 10 + 2 * 3 = 16

So, P + Q = 4 …(ii)

Or, (P – Q)² = P² + Q² – 2PQ = 10 – 2 * 3 = 4.

So, P – Q = 2 …(iii)

Adding (ii) and (iii).

Or, 2P = 6

So, P = 3.

From (i) Q = 4 – 3 = 1

So, P = 3N, Q = 1N.

 

8) Two forces PN and Q N include an angle of 120^o and their resultant is √19 N. If the included angle between the forces were 60 ^o, their resultant would be 7 N. Find P and Q

Solution:

R² = P² + Q² + 2PQ.cosα

Or, 19 = P² + Q² + 2PQcos120°.

So, 19 = P² + Q² – PQ ….(i)

Or, R² = P² + Q² + 2PQcos60°.

Or, 49 = P² + Q² + PQ….(ii).

Adding (i) and (ii),

Or, 68 = 2P² + 2Q².

Or, P² + Q² = 34.

Subtracting (i) and (ii),

Or, 2PQ = 30

So, PQ = 15.

Or, (P + Q)² = P² + Q² + 2PQ = 34 + 2 * 15 = 64.

So, P + Q = 8 ….(iii)

Or, (P – Q)² = P² + Q² – 2PQ = 34 – 2 * 15 = 4

So, P – Q = 2 …(iv)

Adding (iii) and (iv)

Or, 2P = 10, So, P = 5N.

From (iii), Q = 8 – 5 = 3N.

 

9) The sum of the forces is 18 N and their resultant which is perpendicular to the smaller of the two forces is 12 N. Find the magnitude of the forces.

Solution:

P + Q = 18 ….(i)

Or, tan 90° = $\frac{{{\rm{Qsin}}\alpha }}{{{\rm{P}} + {\rm{Qcos}}\alpha }}$

Or, P + Qcosα = 0.

So, Qcosα= -P.

So, cosα = $ - \frac{{\rm{P}}}{{\rm{Q}}}$.

R² = P² + Q² + 2PQ cosα.

Or, 144 = P² + Q² + 2PQ(-P/Q)

Or, 144 = Q² – P²

Or, 144 = (Q + P)(Q – P).

Or, 144 = 18(Q – P)

So, Q – P = 8 ….(ii).

Adding (i) and (ii),

2Q = 26

So, Q = 13.

From(i) P + 13 = 18

So, P = 5.

So, P = 5N, Q = 13N.

 

10) Two forces of magnitude 3P, 2P respectively have a resultant R. If the first force be doubled, the magnitude of the resultant is doubled. Find the angle between the forces.

Solution:

R² = (3P)² + (2P)² + 2.3P.2Pcosα = 9P² + 4P² + 12P².cosα

= 13P² + 12P².cosα …(i)

Again (2R)² = (6P)² + (2P)² + 2.6P.2P.cosα

Or, 4R² = 36P² + 4P² + 24P²cosα.

Or, 4(13P² + 12P².cosα) = 36P² + 4P² + 24P²cosα [From (i)].

Or, 52P² + 48P².cosα = 40P² + 24P².cosα.

Or, 24P².cosα = -12P².

Or, cosα = $ - \frac{1}{2}$.

So, α= 120°.

 

11. The resultant of two forces P and Q is equal to √3Q and making an angle of 30° with the direction of P; show that P is either equal to Q or is double of Q.

Solution:

If OA and OB represent the forces P and Q, then the diagonal OC of the parallelogram OACB represent the resultant $\sqrt 3 $Q.

By given, $\angle $AOC = 30°.

Because AC and OB are equal and parallel so, AC represents Q.

Using cosine law,

Or, cos 30° = $\frac{{{\rm{O}}{{\rm{A}}^2} + {\rm{O}}{{\rm{C}}^2} - {\rm{A}}{{\rm{C}}^2}}}{{2.{\rm{OA}}.{\rm{OC}}}}$

Or, $\frac{{\sqrt 3 }}{2}$ = $\frac{{{{\rm{P}}^2} + 3{{\rm{Q}}^2} - {{\rm{Q}}^2}}}{{2{\rm{p}}.\sqrt 3 {\rm{Q}}}}$.

Or, P² + 2Q² = 3PQ.

Or, P² – 3PQ + 2Q² = 0.

Or, (P – Q)(P – 2Q) = 0

Either, P – Q = 0

SO, P = Q.

Or, P – 2Q = 0

So, P = 2Q.

So, P = Q or 2Q.

 

12. The resultant of two forces P and Q acting at an angle α is equal to (2m+1) $\sqrt {{P^2} + {Q^2}} $. When they act at an angle (90°-α) the resultant is (2m-1$\sqrt {{P^2} + {Q^2}} $ Prove that:

\[\tan \alpha  = \frac{{m - 1}}{{m + 1}}\]

Solution:

${\left\{ {\left( {2{\rm{m}} + 1} \right)\sqrt {{{\rm{P}}^2} + {{\rm{Q}}^2}} } \right\}^2}$ = P² + Q² + 2PQ.cosα

Or, (2m + 1)²(P² + Q²) – (P² + Q²) = 2PQ.cosα.

Or, (P² + Q²){(2m + 1)² – 1} = 2PQ.cosα

Or, (P² + Q²)(4m² + 4m) = 2PQcosα.

Or, (P² + Q²).4m(m + 1) = 2PQ.cosα …(i)

Again,

Or, ${\left\{ {\left( {2{\rm{m}} - 1} \right)\sqrt {{{\rm{P}}^2} + {{\rm{Q}}^2}} } \right\}^2}$ = P² + Q² + 2PQ.cos(90° - α).

Or, (2m – 1)²(P² + Q²) – (P² + Q²) = 2PQsinα.

Or, (P² + Q²){(2m – 1)² – 1} = 2PQ.sinα.

Or, (P² + Q²)(4m² – 4m) = 2PQ.sinα.

Or, (P² + Q²)4m(m – 1) = 2PQsinα …(Ii)

From (i) and (ii)

Or, $\frac{{2{\rm{PQ}}.{\rm{sin}}\alpha }}{{2{\rm{PQcos}}\alpha }}$ = $\frac{{\left( {{{\rm{P}}^2} + {{\rm{Q}}^2}} \right)4{\rm{m}}\left( {{\rm{m}} - 1} \right)}}{{\left( {{{\rm{P}}^2} + {{\rm{Q}}^2}} \right)4{\rm{m}}\left( {{\rm{m}} + 1} \right)}}$

tan α = $\frac{{{\rm{m}} - 1}}{{{\rm{m}} + 1}}$.

 

13. Two forces P and Q acting at a point have a resultant R. If Q be doubled, R is doubled and if Q is reversed in direction, R is again doubled. Show that P : Q : R = √2 : √3 : √2.

Solution:

R² = P² + Q² + 2PQcosα ….(i)

Or, (2R)² = P² +(2Q)² + 2.P.2Q.cosα.

Or, 4R² = P² + 4Q² + 4PQ.cosα ….(ii)

Again, (2R)² = P² + Q² + 2PQ. Cos(π – α).

Or, 4R² = P² + Q² – 2PQcosα ….(iii)

Adding (i) and (iii),

Or, 5R² = 2P² + 2Q²

Or, 2P² + 2Q² – 5R² = 0 ….(iv)

Multiplying (iii) by 2 and then adding with (ii),

Or, 12R² = 3P² + 6Q²

Or, P² + 2Q² – 4R² = 0 ….(v)

From (iv) and (v)

Or, $\frac{{{{\rm{P}}^2}}}{{ - 8 + 10}}$ = $\frac{{{{\rm{Q}}^2}}}{{ - 5 + 8}}$ = $\frac{{{{\rm{R}}^2}}}{{4 - 2}}$.

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

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

So, P:Q:R = $\sqrt 2 :\sqrt 3 :\sqrt 2 $.