Question

59 vectors each of magnitude 12 N are acting at a
point simultaneously such that each vector makes an
angle 6° with the next vector. The resultant of these
vectors is​

Answer 1

Let the first vector be,

$\longrightarrow\vec{\sf{a_1}}=\sf{12\left(\cos6^o\,\hat i+\sin6^o\,\hat j\right)}$

Then second vector can be,

$\longrightarrow\vec{\sf{a_2}}=\sf{12\left(\cos(6+6)^o\,\hat i+\sin(6+6)^o\,\hat j\right)}$

$\longrightarrow\vec{\sf{a_2}}=\sf{12\left(\cos12^o\,\hat i+\sin12^o\,\hat j\right)}$

Likewise, the $\sf{r^{th}}$ vector becomes,

$\longrightarrow\vec{\sf{a_r}}=\sf{12\left(\cos(6r)^o\,\hat i+\sin(6r)^o\,\hat j\right)}$

Let us have $\sf{60^{th}}$ vector too.

$\longrightarrow\vec{\sf{a_{60}}}=\sf{12\left(\cos(6\times60)^o\,\hat i+\sin(6\times60)^o\,\hat j\right)}$

$\longrightarrow\vec{\sf{a_{60}}}=\sf{12\left(\cos360^o\,\hat i+\sin360^o\,\hat j\right)}$

$\longrightarrow\vec{\sf{a_{60}}}=\sf{12\,\hat i}$

Resultant of our 59 vectors is,

$\longrightarrow \vec{\sf{R}}=\mathsf{\displaystyle\sum_{r=1}^{59}}\vec{\sf{a_r}}$

$\longrightarrow\vec{\sf{R}}=\left(\displaystyle\mathsf{\sum_{r=1}^{60}}\vec{\sf{a_r}}\right)-\vec{\sf{a_{60}}}\quad\quad\dots\sf{(1)}$

We see that,

$\longrightarrow\vec{\sf{a}}_{\sf{30+r}}=\sf{12\left(\cos(6(30+r))^o\,\hat i+\sin(6(30+r))^o\,\hat j\right)}$

$\longrightarrow\vec{\sf{a}}_{\sf{30+r}}=\sf{12\left(\cos(180+6r)^o\,\hat i+\sin(180+6r)^o\,\hat j\right)}$

$\longrightarrow\vec{\sf{a}}_{\sf{30+r}}=\sf{12\left(-\cos(6r)^o\,\hat i-\sin(6r)^o\,\hat j\right)}$

$\longrightarrow\vec{\sf{a}}_{\sf{30+r}}=\sf{-12\left(\cos(6r)^o\,\hat i+\sin(6r)^o\,\hat j\right)}$

$\longrightarrow\vec{\sf{a}}_{\sf{30+r}}=-\vec{\sf{a_r}}\quad\quad\dots\sf{(2)}$

Then,

$\displaystyle\longrightarrow\mathsf{\sum_{r=1}^{60}}\vec{\sf{a_r}}=\mathsf{\sum_{r=1}^{30}}\vec{\sf{a_r}}+\mathsf{\sum_{r=31}^{60}}\vec{\sf{a_r}}$

$\displaystyle\longrightarrow\mathsf{\sum_{r=1}^{60}}\vec{\sf{a_r}}=\mathsf{\sum_{r=1}^{30}}\vec{\sf{a_r}}+\mathsf{\sum_{r=1}^{30}}\vec{\sf{a}}_{\sf{30+r}}\quad\quad\!\!\sf{\left[\because\sum_{r=a}^bf(r)=\sum_{r=a-k}^{b-k}f(r+k)\right]}$

From (2),

$\displaystyle\longrightarrow\mathsf{\sum_{r=1}^{60}}\vec{\sf{a_r}}=\mathsf{\sum_{r=1}^{30}}\vec{\sf{a_r}}+\mathsf{\sum_{r=1}^{30}}-\vec{\sf{a_r}}$

$\displaystyle\longrightarrow\mathsf{\sum_{r=1}^{60}}\vec{\sf{a_r}}=\mathsf{\sum_{r=1}^{30}}\vec{\sf{a_r}}-\mathsf{\sum_{r=1}^{30}}\vec{\sf{a_r}}$

$\displaystyle\longrightarrow\mathsf{\sum_{r=1}^{60}}\vec{\sf{a_r}}=\vec{\sf{0}}$

Then (1) becomes,

$\longrightarrow\vec{\sf{R}}=\vec{\sf{0}}-\vec{\sf{a_{60}}}$

$\longrightarrow\vec{\sf{R}}=-\vec{\sf{a_{60}}}$

$\longrightarrow\vec{\sf{R}}=-\vec{\sf{a}}_{\sf{30+30}}$

From (2),

$\longrightarrow\vec{\sf{R}}=-\left(-\vec{\sf{a_{30}}}\right)$

$\longrightarrow\underline{\underline{\vec{\sf{R}}=\vec{\sf{a_{30}}}}}$

Hence 30th vector among them is the resultant.

Answer 2

Answer:

Let the first vector be,

\longrightarrow\vec{\sf{a_1}}=\sf{12(\cos6^o\,\hat i+\sin6^o\,\hat j)}⟶

a

1

=12(cos6

o

i

^

+sin6

o

j

^

)

Then second vector can be,

\longrightarrow\vec{\sf{a_2}}=\sf{12(\cos(6+6)^o\,\hat i+\sin(6+6)^o\,\hat j)}⟶

a

2

=12(cos(6+6)

o

i

^

+sin(6+6)

o

j

^

)

\longrightarrow\vec{\sf{a_2}}=\sf{12(\cos12^o\,\hat i+\sin12^o\,\hat j)}⟶

a

2

=12(cos12

o

i

^

+sin12

o

j

^

)

Likewise, the \sf{r^{th}}r

th

vector becomes,

\longrightarrow\vec{\sf{a_r}}=\sf{12(\cos(6r)^o\,\hat i+\sin(6r)^o\,\hat j)}⟶

a

r

=12(cos(6r)

o

i

^

+sin(6r)

o

j

^

)

Let us have \sf{60^{th}}60

th

vector too.

\longrightarrow\vec{\sf{a_{60}}}=\sf{12(\cos(6\times60)^o\,\hat i+\sin(6\times60)^o\,\hat j)}⟶

a

60

=12(cos(6×60)

o

i

^

+sin(6×60)

o

j

^

)

\longrightarrow\vec{\sf{a_{60}}}=\sf{12(\cos360^o\,\hat i+\sin360^o\,\hat j)}⟶

a

60

=12(cos360

o

i

^

+sin360

o

j

^

)

\longrightarrow\vec{\sf{a_{60}}}=\sf{12\,\hat i}⟶

a

60

=12

i

^

Resultant of our 59 vectors is,

\longrightarrow \vec{\sf{R}}=\mathsf{\displaystyle\sum_{r=1}^{59}}\vec{\sf{a_r}}⟶

R

=

r=1

∑

59

a

r

\longrightarrow\vec{\sf{R}}=(\displaystyle\mathsf{\sum_{r=1}^{60}}\vec{\sf{a_r}})-\vec{\sf{a_{60}}}\quad\quad\dots\sf{(1)}⟶

R

=(

r=1

∑

60

a

r

)−

a

60

…(1)

We see that,

\longrightarrow\vec{\sf{a}}_{\sf{30+r}}=\sf{12(\cos(6(30+r))^o\,\hat i+\sin(6(30+r))^o\,\hat j)}⟶

a

30+r

=12(cos(6(30+r))

o

i

^

+sin(6(30+r))

o

j

^

)

\longrightarrow\vec{\sf{a}}_{\sf{30+r}}=\sf{12(\cos(180+6r)^o\,\hat i+\sin(180+6r)^o\,\hat j)}⟶

a

30+r

=12(cos(180+6r)

o

i

^

+sin(180+6r)

o

j

^

)

\longrightarrow\vec{\sf{a}}_{\sf{30+r}}=\sf{12(-\cos(6r)^o\,\hat i-\sin(6r)^o\,\hat j)}⟶

a

30+r

=12(−cos(6r)

o

i

^

−sin(6r)

o

j

^

)

\longrightarrow\vec{\sf{a}}_{\sf{30+r}}=\sf{-12(\cos(6r)^o\,\hat i+\sin(6r)^o\,\hat j)}⟶

a

30+r

=−12(cos(6r)

o

i

^

+sin(6r)

o

j

^

)

\longrightarrow\vec{\sf{a}}_{\sf{30+r}}=-\vec{\sf{a_r}}\quad\quad\dots\sf{(2)}⟶

a

30+r

=−

a

r

…(2)

Then,

\displaystyle\longrightarrow\mathsf{\sum_{r=1}^{60}}\vec{\sf{a_r}}=\mathsf{\sum_{r=1}^{30}}\vec{\sf{a_r}}+\mathsf{\sum_{r=31}^{60}}\vec{\sf{a_r}}⟶

r=1

∑

60

a

r

=

r=1

∑

30

a

r

+

r=31

∑

60

a

r

\displaystyle\longrightarrow\mathsf{\sum_{r=1}^{60}}\vec{\sf{a_r}}=\mathsf{\sum_{r=1}^{30}}\vec{\sf{a_r}}+\mathsf{\sum_{r=1}^{30}}\vec{\sf{a}}_{\sf{30+r}}\quad\quad\!\!\sf{[\because\sum_{r=a}^bf(r)=\sum_{r=a-k}^{b-k}f(r+k)]}⟶

r=1

∑

60

a

r

=

r=1

∑

30

a

r

+

r=1

∑

30

a

30+r

[∵

r=a

∑

b

f(r)=

r=a−k

∑

b−k

f(r+k)]

From (2),

\displaystyle\longrightarrow\mathsf{\sum_{r=1}^{60}}\vec{\sf{a_r}}=\mathsf{\sum_{r=1}^{30}}\vec{\sf{a_r}}+\mathsf{\sum_{r=1}^{30}}-\vec{\sf{a_r}}⟶

r=1

∑

60

a

r

=

r=1

∑

30

a

r

+

r=1

∑

30

−

a

r

\displaystyle\longrightarrow\mathsf{\sum_{r=1}^{60}}\vec{\sf{a_r}}=\mathsf{\sum_{r=1}^{30}}\vec{\sf{a_r}}-\mathsf{\sum_{r=1}^{30}}\vec{\sf{a_r}}⟶

r=1

∑

60

a

r

=

r=1

∑

30

a

r

−

r=1

∑

30

a

r

\displaystyle\longrightarrow\mathsf{\sum_{r=1}^{60}}\vec{\sf{a_r}}=\vec{\sf{0}}⟶

r=1

∑

60

a

r

=

0

Then (1) becomes,

\longrightarrow\vec{\sf{R}}=\vec{\sf{0}}-\vec{\sf{a_{60}}}⟶

R

=

0

−

a

60

\longrightarrow\vec{\sf{R}}=-\vec{\sf{a_{60}}}⟶

R

=−

a

60

\longrightarrow\vec{\sf{R}}=-\vec{\sf{a}}_{\sf{30+30}}⟶

R

=−

a

30+30

From (2),

\longrightarrow\vec{\sf{R}}=-(-\vec{\sf{a_{30}}})⟶

R

=−(−

a

30

)

\longrightarrow\underline{\underline{\vec{\sf{R}}=\vec{\sf{a_{30}}}}}⟶

R

=

a

30

Hence 30th vector among them is the resultant.