If the maximum resultant of two vector forces is P and the minimum resultant is Q respectively, then what will be the resultant of these forces at right angles?
(a) $ R = \sqrt {{{\left( {\overrightarrow P } \right)}^2} + {{\left( {\overrightarrow Q } \right)}^2}} $
(b) $ R = \sqrt {\dfrac{{{{\left( {\overrightarrow P } \right)}^2} + {{\left( {\overrightarrow Q } \right)}^2}}}{2}} $
(c) $ R = \sqrt {\dfrac{{{{\left( {\overrightarrow P } \right)}^2} - {{\left( {\overrightarrow Q } \right)}^2}}}{2}} $
(d) Cannot be determined
Answer
572.7k+ views
Hint: We will use the most eccentric concept of vectors. Assuming the forces (as per as the parameters asked in the respective question is concerned). Taking summation of the forces at vectors $ \overrightarrow P $ and $ \overrightarrow Q $ respectively, solving the equation mathematically the required resultant can be obtained.
Complete step-by-step answer:
Since, we have given the two resultant vector forces ‘P’ and ‘Q’ respectively,
By the definition of the vector quantity, the magnitude and direction at right angles may be distinguish by assuming the following parameters,
Let us assume that, ‘ $ {F_1} $ ’ and ‘ $ {F_2} $ ’ are the required resultant forces for vectors ‘ $ \overrightarrow P $ ’ and ‘ $ \overrightarrow Q $ ’ respectively,
As a result, summating both the forces for $ \overrightarrow P $ and $ \overrightarrow Q $ respectively, we get
Hence, Maximum resultant at vector ‘P’ is,
$ \overrightarrow P = {F_1} + {F_2} $ … (i)
Similarly, Minimum resultant at vector ‘Q’ is,
$ \overrightarrow Q = {F_1} - {F_2} $ … (ii)
Adding equations (i) and (ii) that is to find $ {F_1} $ , we get
$
\overrightarrow P + \overrightarrow Q = 2{F_1} \\
\therefore {F_1} = \dfrac{{\overrightarrow P + \overrightarrow Q }}{2} \\
$
Similarly, subtracting equations (i) and (ii) that is to find $ {F_2} $ , we get
$
\overrightarrow P - \overrightarrow Q = 2{F_2} \\
\therefore {F_2} = \dfrac{{\overrightarrow P - \overrightarrow Q }}{2} \\
$
Now, by the definition of resultant (of its magnitude), we get
\[{\text{Resultant}} = R = \sqrt {F_1^2 + F_2^2} \]
Substituting the values of $ {F_1} $ and $ {F_2} $ respectively, we get
\[
R = \sqrt {{{\left( {\dfrac{{\overrightarrow P + \overrightarrow Q }}{2}} \right)}^2} + {{\left( {\dfrac{{\overrightarrow P - \overrightarrow Q }}{2}} \right)}^2}} \\
R = \sqrt {\dfrac{1}{4}\left[ {{{\left( {\overrightarrow P + \overrightarrow Q } \right)}^2} + {{\left( {\overrightarrow P - \overrightarrow Q } \right)}^2}} \right] } \\
\]
(Here, we have used an algebraic identity i.e. $ {(a + b)^2} = {a^2} + 2ab + {b^2} $ respectively!
\[\therefore R = \sqrt {\dfrac{1}{4}\left\{ {\left[ {{{\left( {\overrightarrow P } \right)}^2} + 2\overrightarrow {PQ} + {{\left( {\overrightarrow Q } \right)}^2}} \right] + {{\left( {\overrightarrow P } \right)}^2} - 2\overrightarrow {PQ} + {{\left( {\overrightarrow Q } \right)}^2}} \right\}} \]
Solving the equations predominantly, we get
\[
R = \sqrt {\dfrac{1}{4}\left[ {{{\left( {\overrightarrow P } \right)}^2} + 2\overrightarrow {PQ} + {{\left( {\overrightarrow Q } \right)}^2} + {{\left( {\overrightarrow P } \right)}^2} - 2\overrightarrow {PQ} + {{\left( {\overrightarrow Q } \right)}^2}} \right] } \\
R = \sqrt {\dfrac{1}{4}\left[ {{{\left( {\overrightarrow P } \right)}^2} + {{\left( {\overrightarrow Q } \right)}^2} + {{\left( {\overrightarrow P } \right)}^2} + {{\left( {\overrightarrow Q } \right)}^2}} \right] } \\
\]
As a result, adding the same terms (same vectors), we get
\[
R = \sqrt {\dfrac{1}{4}\left[ {2{{\left( {\overrightarrow P } \right)}^2} + 2{{\left( {\overrightarrow Q } \right)}^2}} \right] } \\
R = \sqrt {\dfrac{1}{2}\left[ {{{\left( {\overrightarrow P } \right)}^2} + {{\left( {\overrightarrow Q } \right)}^2}} \right] } \\
\]
Hence, the required resultant of the forces is,
\[R = \sqrt {\dfrac{{{{\left( {\overrightarrow P } \right)}^2} + {{\left( {\overrightarrow Q } \right)}^2}}}{2}} \]
Therefore, option (b) is correct!
So, the correct answer is “Option b”.
Note: One must remember the concept of vector quantity that revolves in magnitude as well as direction of the quantities or applications of the respective parameters. Should clarify how to calculate the resultant of the required forces for the vector quantity particularly that is \[{\text{Resultant}} = \sqrt {F_1^2 + F_2^2} \] . Algebraic identities play a significant role in solving this problem so as for the ease of such problems.
Complete step-by-step answer:
Since, we have given the two resultant vector forces ‘P’ and ‘Q’ respectively,
By the definition of the vector quantity, the magnitude and direction at right angles may be distinguish by assuming the following parameters,
Let us assume that, ‘ $ {F_1} $ ’ and ‘ $ {F_2} $ ’ are the required resultant forces for vectors ‘ $ \overrightarrow P $ ’ and ‘ $ \overrightarrow Q $ ’ respectively,
As a result, summating both the forces for $ \overrightarrow P $ and $ \overrightarrow Q $ respectively, we get
Hence, Maximum resultant at vector ‘P’ is,
$ \overrightarrow P = {F_1} + {F_2} $ … (i)
Similarly, Minimum resultant at vector ‘Q’ is,
$ \overrightarrow Q = {F_1} - {F_2} $ … (ii)
Adding equations (i) and (ii) that is to find $ {F_1} $ , we get
$
\overrightarrow P + \overrightarrow Q = 2{F_1} \\
\therefore {F_1} = \dfrac{{\overrightarrow P + \overrightarrow Q }}{2} \\
$
Similarly, subtracting equations (i) and (ii) that is to find $ {F_2} $ , we get
$
\overrightarrow P - \overrightarrow Q = 2{F_2} \\
\therefore {F_2} = \dfrac{{\overrightarrow P - \overrightarrow Q }}{2} \\
$
Now, by the definition of resultant (of its magnitude), we get
\[{\text{Resultant}} = R = \sqrt {F_1^2 + F_2^2} \]
Substituting the values of $ {F_1} $ and $ {F_2} $ respectively, we get
\[
R = \sqrt {{{\left( {\dfrac{{\overrightarrow P + \overrightarrow Q }}{2}} \right)}^2} + {{\left( {\dfrac{{\overrightarrow P - \overrightarrow Q }}{2}} \right)}^2}} \\
R = \sqrt {\dfrac{1}{4}\left[ {{{\left( {\overrightarrow P + \overrightarrow Q } \right)}^2} + {{\left( {\overrightarrow P - \overrightarrow Q } \right)}^2}} \right] } \\
\]
(Here, we have used an algebraic identity i.e. $ {(a + b)^2} = {a^2} + 2ab + {b^2} $ respectively!
\[\therefore R = \sqrt {\dfrac{1}{4}\left\{ {\left[ {{{\left( {\overrightarrow P } \right)}^2} + 2\overrightarrow {PQ} + {{\left( {\overrightarrow Q } \right)}^2}} \right] + {{\left( {\overrightarrow P } \right)}^2} - 2\overrightarrow {PQ} + {{\left( {\overrightarrow Q } \right)}^2}} \right\}} \]
Solving the equations predominantly, we get
\[
R = \sqrt {\dfrac{1}{4}\left[ {{{\left( {\overrightarrow P } \right)}^2} + 2\overrightarrow {PQ} + {{\left( {\overrightarrow Q } \right)}^2} + {{\left( {\overrightarrow P } \right)}^2} - 2\overrightarrow {PQ} + {{\left( {\overrightarrow Q } \right)}^2}} \right] } \\
R = \sqrt {\dfrac{1}{4}\left[ {{{\left( {\overrightarrow P } \right)}^2} + {{\left( {\overrightarrow Q } \right)}^2} + {{\left( {\overrightarrow P } \right)}^2} + {{\left( {\overrightarrow Q } \right)}^2}} \right] } \\
\]
As a result, adding the same terms (same vectors), we get
\[
R = \sqrt {\dfrac{1}{4}\left[ {2{{\left( {\overrightarrow P } \right)}^2} + 2{{\left( {\overrightarrow Q } \right)}^2}} \right] } \\
R = \sqrt {\dfrac{1}{2}\left[ {{{\left( {\overrightarrow P } \right)}^2} + {{\left( {\overrightarrow Q } \right)}^2}} \right] } \\
\]
Hence, the required resultant of the forces is,
\[R = \sqrt {\dfrac{{{{\left( {\overrightarrow P } \right)}^2} + {{\left( {\overrightarrow Q } \right)}^2}}}{2}} \]
Therefore, option (b) is correct!
So, the correct answer is “Option b”.
Note: One must remember the concept of vector quantity that revolves in magnitude as well as direction of the quantities or applications of the respective parameters. Should clarify how to calculate the resultant of the required forces for the vector quantity particularly that is \[{\text{Resultant}} = \sqrt {F_1^2 + F_2^2} \] . Algebraic identities play a significant role in solving this problem so as for the ease of such problems.
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