Exercise 21.2
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
F = 10N, α= 30°.
P = Fcosα = 10.cos30° = 10 *
$\frac{{\sqrt 3 }}{2}$ = 5$\sqrt 3 $N.
Q = F.sinα = 10.sin30° = 10 *
$\frac{1}{2}$ = 5N.
2) Resolve a force equal to 50 N into two forces making angles of 60°
and 45° with it an opposite sides.
Solution
R = 50N, α= 60°, β = 45°.
P = $\frac{{{\rm{Rsin}}\beta
}}{{\sin \left( {\alpha + \beta } \right)}}$ =
$\frac{{50{\rm{sin}}45\infty }}{{{\rm{sin}}105\infty }}$ =
$\frac{{50{\rm{*}}\frac{1}{{\sqrt 2 }}}}{{\frac{{\sqrt 3 + 1}}{{2\sqrt 2
}}}}$ = $\frac{{100}}{{\sqrt 3 + 1}}$ * $\frac{{\sqrt 3 -
1}}{{\sqrt 3 - 1}}$ = 50$\left( {\sqrt 3 - 1} \right)$N.
Q = $\frac{{{\rm{Rsin}}\alpha
}}{{\sin \left( {\alpha + \beta } \right)}}$ =
$\frac{{50{\rm{sin}}60\infty }}{{{\rm{sin}}105\infty }}$ =
$\frac{{50{\rm{*}}\frac{{\sqrt 3 }}{2}}}{{\frac{{\sqrt 3 + 1}}{{2\sqrt 2
}}}}$ = $\frac{{50\sqrt 6 }}{{\sqrt 3 + 1}}$ = $\frac{{50\sqrt 6 \left(
{\sqrt 3 - 1} \right)}}{2}$ = 25$\left( {3\sqrt 2 - \sqrt 6 }
\right)$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 $\frac{{\sqrt {6} }}{3}$P.
Solution
R = P, α= 45°, β = 15°.
Q = $\frac{{{\rm{Rsin}}\alpha
}}{{\sin \left( {\alpha + \beta } \right)}}$ =
$\frac{{{\rm{Psin}}45\infty }}{{{\rm{sin}}60\infty }}$ =
$\frac{{{\rm{P*}}\frac{1}{{\sqrt 2 }}}}{{\frac{{\sqrt 3 }}{2}}}$ =
$\frac{{2{\rm{P}}}}{{\sqrt 6 }}$ = $\frac{{2\sqrt 6 }}{{\sqrt 6 }}$P =
$\frac{{\sqrt 6 }}{3}$P.
4) A force of 20 N acting vertically upwards is resolved into two
forces one being horizontal and equal to 10 N. What is the magnitude and
direction of the other force?
Solution
P = 10N, R = 20N, θ= 90°, Q = ?.
tanθ = $\frac{{{\rm{Qsin}}\alpha
}}{{{\rm{P}} + {\rm{Q}}.{\rm{sin}}\alpha }}$
tan90° =
$\frac{{{\rm{Qsin}}\alpha }}{{10 + {\rm{Qcos}}\alpha }}$.
∞ = $\frac{{{\rm{Qsin}}\alpha
}}{{10 + {\rm{Qcos}}\alpha }}$.
So, cosα = $ -
\frac{{10}}{{\rm{Q}}}$.
R2 = P2 +
Q2 + 2PQcosα.
Or, (20)2 = (10)2 +
Q2 + 2.10.Q$\left( { - \frac{{10}}{{\rm{Q}}}} \right)$.
Or, 400 = 100 + Q2 –
200
Or, Q2 = 500,
So, Q = 10$\sqrt 5 $N.
If β be the angle made by Q with
vertical line:
Or, R2 = P2 +
Q2 + 2PQcos(90° + β).
Or, 400 = 100 + 500 +
2.10.10$\sqrt 5 $ (-sinβ)
Or, - 200 = - 200$\sqrt 5 $sinβ.
So, sinβ = $\frac{1}{{\sqrt 5
}}$.
Cosβ = $\sqrt {1 - \frac{1}{5}} $
= $\frac{2}{{\sqrt 5 }}$.
Tanβ = $\frac{{\frac{1}{{\sqrt 5
}}}}{{2\sqrt 5 }}$ = $\frac{1}{2}$.
So, β = tan-1$\left(
{\frac{1}{2}} \right)$.
5) A particle is acted upon by three forces, in one plane, equal to 2,
2√2 and 1 newton respectively. The first is horizontal, the second acts at 45°
to the horizon, and the third is vertical. Find their resultant.
Solution
Resolving the forces along and
perpendicular to OA,
Or, X = 2cos0° + 2$\sqrt 2 $
cos45° + 1.cos90°
= 2.1 + 2$\sqrt 2 $.$\frac{1}{{\sqrt
2 }}$ + 0 = 4.
Y = 2sin0° + 2$\sqrt 2 $ sin45° +
1.sin90°.
= 0 + 2$\sqrt 2 $.
$\frac{1}{{\sqrt 2 }}{\rm{\: }}$ + 1 = 3.
R = $\sqrt {{{\rm{X}}^2} +
{{\rm{Y}}^2}} $ = $\sqrt {16 + 9} $ = 5N.
Tanθ =
$\frac{{\rm{Y}}}{{\rm{X}}}$ = $\frac{3}{4}$.
So, θ = tan-1$\left(
{\frac{3}{4}} \right)$.
6. Forces 1 N, 2 N and 3 N act at a point in direction parallel to the
side of an equilateral triangle taken in order. Find their resultant.
Solution
Resolving forces along and
perpendicular to BC,
X = 1.cos0° + 2.cos120° +
3.cos240°.
= 1 + 2.$\left( { - \frac{1}{2}}
\right)$ + 3.$\left( { - \frac{1}{2}} \right)$ = 1 – 1 – $\frac{3}{2}$ = $ -
\frac{3}{2}$.
Y = 1.sin0° + 2.sin120° +
3.cos240°.
= 0 + 2.$\frac{{\sqrt 3 }}{2}$ +
3.$\left( { - \frac{{\sqrt 3 }}{2}} \right)$ = $ - \frac{{\sqrt 3 }}{2}$.
R = $\sqrt {{{\rm{X}}^2} +
{{\rm{Y}}^2}} $ = $\sqrt {\frac{9}{4} + \frac{3}{4}} $ = $\sqrt 3 $N.
Tanθ = $\frac{{\rm{Y}}}{{\rm{X}}}$
= $\frac{{ - \frac{{\sqrt 3 }}{2}}}{{ - \frac{3}{2}}}$ = $\frac{1}{{\sqrt 3
}}$.
So, θ = 180° + 30° = 210°.
The angle made by the resultant
second force = 210° - 120°= 90°.
7. ABCD is a square. Forces 1, 2, 3, 4 and 2√2 newton act at a point in
directions AB, BC. CD, DA and AC. Find the resultant.
Solution
Resolving the forces along and
perpendicular to AB,
X = 1cos0° +2.cos90° + 3cos180° +
4cos270° + 2$\sqrt 2 $cos45°.
= 1.1 + 0 + 3.(-1) + 4.0 +
2$\sqrt 2 $ .$\left( {\frac{1}{{\sqrt 2 }}} \right)$ = 1 – 3 + 2 = 0
Y = 1sin0° + 2sin90° + 3sin180° +
4sin270° + $2\sqrt 2 $sin45°.
= 0 + 2.1 + 3.0 + 4.(-1) +
2$\sqrt 2 $ .$\frac{1}{{\sqrt 2 }}$ = 2 – 4 + 2 = 0
R = $\sqrt {{{\rm{X}}^2} + {{\rm{Y}}^2}}
$ = 0
So, forces are in equilibrium.
8. Forces of 2, √3, 5, √3 and 2 newton respectively act at one of the
angular points of a regular hexagon towards the five other points. Find the magnitude
and direction of the resultant.
Solution
Resolving the forces along and
perpendicular to AB.
X = 2cos0° + $\sqrt 3 $cos30° + 5.cos60° + $\sqrt 3 $cos90° + 2.cos120°.
= 2.1 + $\sqrt 3 $.$\frac{{\sqrt
3 }}{2}$ + 5.$\frac{1}{2}$ + $\sqrt 3 $.0 + 2.$\left( { - \frac{1}{2}}
\right)$.
= 2 + $\frac{3}{2}$ +
$\frac{5}{2}$ – 1 = 5.
Y = 2sin0° + $\sqrt 3 $sin30° +
5sin60° + $\sqrt 3 $sin90° + 2sin120°.
= 0 + $\sqrt 3 $.$\frac{1}{2}$ +
5.$\frac{{\sqrt 3 }}{2}$ + $\sqrt 3 $.1 + 2.$\frac{{\sqrt 3 }}{2}$ = 4$\sqrt 3
$.
R = $\sqrt {{{\rm{X}}^2} +
{{\rm{Y}}^2}} $ = $\sqrt {25 + 75} $ = 10N
Tanθ =
$\frac{{\rm{Y}}}{{\rm{X}}}$ = $\frac{{5\sqrt 3 }}{2}$ = $\sqrt 3 $.
θ = 60°.
Thus, R = 10N towards opposite angular points.