
If a body is in equilibrium under the effect of some non-collinear forces, then the minimum number of such forces acting upon the body are
A. \[3\]
B. \[2\]
C. \[5\]
D. \[4\]
Answer
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Hint: It is given that a body is in equilibrium under the effect of some non-collinear forces. We have to find out the minimum number of forces acting on the body to keep it in equilibrium.
Complete step by step answer:
When two forces in two different planes act on a body, we can say that these forces are non-collinear. We will use the equilibrium concept by assuming a rectangular body loaded with non-collinear forces.
Let us consider a rectangular body is loaded with forces \[{F_1}\] and \[{F_2}\] in such a way that \[{F_1}\] is acting in one plane and \[{F_2}\] is acting in another plane.
We know that for a body to be in an equilibrium state, the summation of all the forces acting on it must be zero. We can say that all the forces acting on the body are balanced by the individual force of the same magnitude but opposite in direction, as shown in the figure below.
From the concept of equilibrium of forces, we know that the summation of all the forces is equal to zero.
\[\sum F = 0\]
Rewrite the above expression for x-direction.
\[\sum {{F_x} = 0} \]
Let us take the above expression's right-hand side and substitute the forces acting in the x-direction.
$
{F_1} + \left( { - {F_1}} \right) = {F_1} - {F_1}\\
= 0
$……(1)
Write the expression for the equilibrium of forces in the y-direction.
\[\sum {{F_y} = 0} \]
Let us take the above expression's right-hand side and substitute the forces acting in the y-direction.
$
{F_2} + \left( { - {F_2}} \right) = {F_2} - {F_2}\\
= 0
$……(2)
Therefore, we can say that if a body is in equilibrium under the effect of some non-collinear forces, then the minimum number of such forces acting upon the body will be four.
So, the correct answer is “Option D”.
Note:
We would have applied more forces on the rectangular body, but we are asked to evaluate the minimum number of forces acting to keep the body in equilibrium; that's why only two forces are considered to be in action.
Complete step by step answer:
When two forces in two different planes act on a body, we can say that these forces are non-collinear. We will use the equilibrium concept by assuming a rectangular body loaded with non-collinear forces.
Let us consider a rectangular body is loaded with forces \[{F_1}\] and \[{F_2}\] in such a way that \[{F_1}\] is acting in one plane and \[{F_2}\] is acting in another plane.
We know that for a body to be in an equilibrium state, the summation of all the forces acting on it must be zero. We can say that all the forces acting on the body are balanced by the individual force of the same magnitude but opposite in direction, as shown in the figure below.
From the concept of equilibrium of forces, we know that the summation of all the forces is equal to zero.
\[\sum F = 0\]
Rewrite the above expression for x-direction.
\[\sum {{F_x} = 0} \]
Let us take the above expression's right-hand side and substitute the forces acting in the x-direction.
$
{F_1} + \left( { - {F_1}} \right) = {F_1} - {F_1}\\
= 0
$……(1)
Write the expression for the equilibrium of forces in the y-direction.
\[\sum {{F_y} = 0} \]
Let us take the above expression's right-hand side and substitute the forces acting in the y-direction.
$
{F_2} + \left( { - {F_2}} \right) = {F_2} - {F_2}\\
= 0
$……(2)
Therefore, we can say that if a body is in equilibrium under the effect of some non-collinear forces, then the minimum number of such forces acting upon the body will be four.
So, the correct answer is “Option D”.
Note:
We would have applied more forces on the rectangular body, but we are asked to evaluate the minimum number of forces acting to keep the body in equilibrium; that's why only two forces are considered to be in action.
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