CBSE-XI-Physics
02: Physics and Mathematics
- #3Add vectors
A→,B→andC→each having magnitude of 100 unit and inclined to the X-axis at angles 45°, 135° and 315° respectively.Ans : First, we will find the components of the vector along the x-axis and y-axis. Then we will find the resultant x and y-components.
x-component of `` \stackrel{\to }{\mathrm{A}}=\mathrm{Acos}45°=100\mathrm{cos}45°=\frac{100}{\sqrt{2}}\mathrm{unit}``
x-component of `` \stackrel{\to }{\mathrm{B}}=\stackrel{\to }{\mathrm{B}}\mathrm{cos}135°=-\frac{100}{\sqrt{2}}``
x-component of `` \stackrel{\to }{\mathrm{C}}`` = `` \stackrel{\to }{\mathrm{C}}``cos315`` °`` = 100 cos 315°
`` =100\mathrm{cos}45°=\frac{100}{\sqrt{2}}``
Resultant x-component `` =\frac{100}{\sqrt{2}}-\frac{100}{\sqrt{2}}+\frac{100}{\sqrt{2}}=\frac{100}{\sqrt{2}}``
Now, y-component of `` \stackrel{\to }{\mathrm{A}}=100\mathrm{sin}45°=\frac{100}{\sqrt{2}}``
y-component of `` \stackrel{\to }{\mathrm{B}}=100\mathrm{sin}135°=\frac{100}{\sqrt{2}}``
y-component of `` \stackrel{\to }{\mathrm{C}}=100\mathrm{sin}315°=-\frac{100}{\sqrt{2}}``
Resultant y-component`` =\frac{100}{\sqrt{2}}+\frac{100}{\sqrt{2}}-\frac{100}{\sqrt{2}}=\frac{100}{\sqrt{2}}``
Magnitude of the resultant`` =\sqrt{{\left(\frac{100}{\sqrt{2}}\right)}^{2}+{\left(\frac{100}{\sqrt{2}}\right)}^{2}}``
`` =\sqrt{10000}=100``
Angle made by the resultant vector with the x-axis is given by
`` \mathrm{tan}\mathrm{\alpha }=\frac{y\mathrm{comp}}{x\mathrm{comp}}
=\frac{100\sqrt{2}}{100\sqrt{2}}=1``
⇒ α = tan-1 (1) = 45°
∴ The magnitude of the resultant vector is 100 units and it makes an angle of 45° with the x-axis.
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