Two forces having magnitude A and frac{A}{2} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the two forces be $\vec{F}_1$ and $\vec{F}_2$,where $|\vec{F}_1| = A$ and $|\vec{F}_2| = \frac{A}{2}$.Since the forces are perpendicular to ea

Vedclass · Accepted Answer

$\frac{\sqrt{5}A}{2}$

Vedclass · Answer

(D) Let the two forces be $\vec{F}_1$ and $\vec{F}_2$,where $|\vec{F}_1| = A$ and $|\vec{F}_2| = \frac{A}{2}$.Since the forces are perpendicular to each other,the angle between them is $	heta = 90^{\circ}$.The magnitude of the resultant force $\vec{F} = \vec{F}_1 + \vec{F}_2$ is given by:$|\vec{F}| = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos 90^{\circ}}$Since $\cos 90^{\circ} = 0$,the expression simplifies to:$|\vec{F}| = \sqrt{F_1^2 + F_2^2}$Substituting the given values:$|\vec{F}| = \sqrt{A^2 + \left(\frac{A}{2}ight)^2} = \sqrt{A^2 + \frac{A^2}{4}}$$|\vec{F}| = \sqrt{\frac{4A^2 + A^2}{4}} = \sqrt{\frac{5A^2}{4}}$$|\vec{F}| = \frac{\sqrt{5}A}{2}$

Two forces having magnitude $A$ and $\frac{A}{2}$ are perpendicular to each other. The magnitude of their resultant is

Similar Questions

Write two properties of vector addition.

If $\overrightarrow A = 2\hat i + \hat j$,$\overrightarrow B = 3\hat j - \hat k$,and $\overrightarrow C = 6\hat i - 2\hat k$,find the value of $\overrightarrow A - 2\overrightarrow B + 3\overrightarrow C$.

Two vectors of magnitude $3$ and $4$ have a resultant which makes angles $\alpha$ and $\beta$ respectively with them (given $\alpha + \beta \neq 90^o$).

If $|{\overrightarrow V _1} + {\overrightarrow V _2}| = |{\overrightarrow V _1} - {\overrightarrow V _2}|$ and ${V_2}$ is finite,then

If $P + Q = R$ and $|P| = |Q| = \sqrt{3}$ and $|R| = 3$,then the angle between $P$ and $Q$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module