Name: Lami's Theorem in Equilibrium
Uploaded: 2019-07-19
Duration: 7 min 28 s
Description: In this interesting video, we will learn about Lami's theorem in equilibrium. Watch the solved sample problems to quickly understand the theorem.

Question

What do you mean by resolution of vectors? Explain the resolution of vectors into rectangular components.

Accepted Answer

Resolution of vector:
It is defined as the splitting of a single vector into two or more vectors in different directions which together produce a similar effect as it produced by a single vector itself. And the vector which formed after splitting is known as the component vectors.

Two component of resolved vectors are:
1) Horizontal Component
2) Vertical Component
Horizontal component:
It is that component of the vector that moves directly in a line parallel to the horizontal axis.
Vertical Component:
It is that component of the vector that lies perpendicular to a horizontal or level plane.
Now with the help of diagram we explain the rectangular components in plane

Let us assume a vector $\vec{v}$ which resolves into two components $\vec{v_{x}}$ and $\vec{v_{y}}$ .

Where,
$O B = \vec{v_{x}}$ = Vector along the x-axis
Vector along x-axis is the horizontal component
$O A = \vec{v_{y}}$ = Vector along the y-axis
Vector along y-axis is the vertical component

From this we obtained our horizontal and vertical component of the vector $\vec{v}$ .
Now,
By drawing a line parallel to OA we will get BP which is the same as that of OA as it resolves the vector in rectangular components.
Hence,

$O A = B P = \vec{v_{y}}$

By applying the triangle law of addition in $Δ O B P$ we will get,

$O P = O B + B P$

By putting the components we will get,

$\vec{v} = \vec{v_{x}} + \vec{v_{y}} \dots . . . . . \dots (1)$

Here we can see the $Δ O B P$ is a right-angled triangle, so we can use the trigonometric function to get the angular component along the x and y-axis as well as the magnitude of the vector, respectively.

In $Δ O B P$ ,

Here $θ$ is the angle between OP and OB.

$\cos θ = \frac{B a s e}{H y p o t e n u s e} = \frac{O B}{O P}$

$\Rightarrow O P \cos θ = O B$

Putting the component is pace of OP and OB,

$\vec{v} \cos θ = \vec{v_{x}} \dots . . . . . . . \dots (2)$

Similarly,

$\sin θ = \frac{H e i g h t}{H y p o t e n u s e} = \frac{B P}{O P}$

$\Rightarrow O P \sin θ = B P$

Putting the component is pace of OP and BP,

$\vec{v} \sin θ = \vec{v_{y}} \dots . . . . . . . \dots (3)$

Dividing equation (3) by (2) we will get,

$\frac{\vec{v} \sin θ}{\vec{v} \cos θ} = \frac{\vec{v_{y}}}{\vec{v_{x}}}$

Cancelling the common terms we will get,

$\frac{\sin θ}{\cos θ} = \frac{\vec{v_{y}}}{\vec{v_{x}}} \Rightarrow \tan θ = \frac{\vec{v_{y}}}{\vec{v_{x}}}$

Squaring and adding equation (2) and (3) we will get,

${(\vec{v} \cos θ)}^{2} + {(\vec{v} \sin θ)}^{2} = {(v_{y})}^{2} + {(v_{x})}^{2} \Rightarrow {(\vec{v})}^{2} \cos^{2} θ + {(\vec{v})}^{2} \sin^{2} θ = {(v_{y})}^{2} + {(v_{x})}^{2} \Rightarrow {(\vec{v})}^{2} (\sin^{2} θ + \cos^{2} θ) = {(v_{y})}^{2} + {(v_{x})}^{2} \Rightarrow {(\vec{v})}^{2} = {(v_{y})}^{2} + {(v_{x})}^{2}$

We know,

$\cos^{2} θ + \sin^{2} θ = 1$

Now on further solving we will get,

$\vec{v} = \sqrt{{v_{x}}^{2} + {v_{y}}^{2}}$

What do you mean by resolution of vectors? Explain the resolution of vectors into rectangular components.

Important Questions on Scalars and Vectors

Learn and Prepare for any exam you want !