The adjacent figure is the part of the horizontally stretched net section AB is stretched with a force n of 10N the tension in the section BC and BF are
A) 10N, 11N
B) 10N, 6N
C) 10N, 10N
D) Cant calculate due to insufficient
Answer
638.1k+ views
Hint
In such a diagram, to find the force or tension we can solve such a question using lamis theorem.
$\dfrac{{\;{T_1}}}{{\operatorname{sin} 120}} = \dfrac{{{T_2}}}{{\operatorname{sin} 120}} = \dfrac{{\;{T_{}}}}{{\operatorname{sin} 120}}\sum {y = 0} 1N = kgm\;{s^{ - 2}}$
Where α, β and γ are the angle directly opposite the force vector.
Complete step-by-step answer:
As we have to find the tension of the section BC and BF. It can be easily found using the concept of the Lamis theorem. Lamis theorem-” It is an equation relating the magnitude of the three coplanar concurrent and non collinear vectors, which keeps an object, is the static equilibrium, with the angle directly opposite to the corresponding vectors.”
$\dfrac{{\;{F_1}}}{{\operatorname{sin} \alpha }} = \dfrac{{\;{F_2}}}{{\operatorname{sin} \beta }} = \dfrac{{\;{F_3}}}{{\operatorname{sin} \gamma }}$
Where α, β and γ are the angle directly opposite the force vector.
So simply putting the value in the equation-
$\dfrac{{\;AB}}{{\operatorname{sin} 120}} = \dfrac{{BC}}{{\operatorname{sin} 120}} = \dfrac{{\;BF}}{{\operatorname{sin} 120}}$
Or it can be also written as-
$\dfrac{{\;{T_1}}}{{\operatorname{sin} 120}} = \dfrac{{{T_2}}}{{\operatorname{sin} 120}} = \dfrac{{\;{T_{}}}}{{\operatorname{sin} 120}}$
Hence $T_1 = T_2 = T = 10N$.
Hence all the forces are equal in magnitude so the correct answer is $10N, 10N$.
(C) is the correct option.
Note
It can also be solved using the alternative method like
Considering the body in equilibrium and resolving the forces into the component. So on using
$\sum {x = 0} $
$\sum {y = 0} $
Hence using this we can easily find the solution. The SI unit of the force is Newton. Where 1 Newton is equal to the kg meter per Second Square or numerically it can be also written as
$1N = kgm\;{s^{ - 2}}$.
In such a diagram, to find the force or tension we can solve such a question using lamis theorem.
$\dfrac{{\;{T_1}}}{{\operatorname{sin} 120}} = \dfrac{{{T_2}}}{{\operatorname{sin} 120}} = \dfrac{{\;{T_{}}}}{{\operatorname{sin} 120}}\sum {y = 0} 1N = kgm\;{s^{ - 2}}$
Where α, β and γ are the angle directly opposite the force vector.
Complete step-by-step answer:
As we have to find the tension of the section BC and BF. It can be easily found using the concept of the Lamis theorem. Lamis theorem-” It is an equation relating the magnitude of the three coplanar concurrent and non collinear vectors, which keeps an object, is the static equilibrium, with the angle directly opposite to the corresponding vectors.”
$\dfrac{{\;{F_1}}}{{\operatorname{sin} \alpha }} = \dfrac{{\;{F_2}}}{{\operatorname{sin} \beta }} = \dfrac{{\;{F_3}}}{{\operatorname{sin} \gamma }}$
Where α, β and γ are the angle directly opposite the force vector.
So simply putting the value in the equation-
$\dfrac{{\;AB}}{{\operatorname{sin} 120}} = \dfrac{{BC}}{{\operatorname{sin} 120}} = \dfrac{{\;BF}}{{\operatorname{sin} 120}}$
Or it can be also written as-
$\dfrac{{\;{T_1}}}{{\operatorname{sin} 120}} = \dfrac{{{T_2}}}{{\operatorname{sin} 120}} = \dfrac{{\;{T_{}}}}{{\operatorname{sin} 120}}$
Hence $T_1 = T_2 = T = 10N$.
Hence all the forces are equal in magnitude so the correct answer is $10N, 10N$.
(C) is the correct option.
Note
It can also be solved using the alternative method like
Considering the body in equilibrium and resolving the forces into the component. So on using
$\sum {x = 0} $
$\sum {y = 0} $
Hence using this we can easily find the solution. The SI unit of the force is Newton. Where 1 Newton is equal to the kg meter per Second Square or numerically it can be also written as
$1N = kgm\;{s^{ - 2}}$.
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