Two forces of magnitude P and Q acting at a p | vedclass

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

Question

(B) Let the angle between the forces $P$ and $Q$ be $	heta$.The resultant $R$ is given by the vector sum $\vec{R} = \vec{P} + \vec{Q}$.The resolved p

Vedclass · Accepted Answer

$2{\sin ^{ - 1}}\left( {\frac{P}{{2Q}}} \right)^{\frac{1}{2}}$

Vedclass · Answer

(B) Let the angle between the forces $P$ and $Q$ be $	heta$.The resultant $R$ is given by the vector sum $\vec{R} = \vec{P} + \vec{Q}$.The resolved part of $R$ in the direction of $P$ is given by $P + Q \cos 	heta$.According to the problem,this resolved part is equal to $Q$.Therefore,$P + Q \cos 	heta = Q$.Rearranging the terms,we get $Q \cos 	heta = Q - P$.Thus,$\cos 	heta = \frac{Q - P}{Q} = 1 - \frac{P}{Q}$.Using the trigonometric identity $\cos 	heta = 1 - 2 \sin^2 \left(\frac{	heta}{2}ight)$,we have:$1 - 2 \sin^2 \left(\frac{	heta}{2}ight) = 1 - \frac{P}{Q}$.Subtracting $1$ from both sides,we get $-2 \sin^2 \left(\frac{	heta}{2}ight) = -\frac{P}{Q}$.Therefore,$\sin^2 \left(\frac{	heta}{2}ight) = \frac{P}{2Q}$.Taking the square root,we get $\sin \left(\frac{	heta}{2}ight) = \left(\frac{P}{2Q}ight)^{1/2}$.Finally,the angle $	heta = 2 \sin^{-1} \left(\frac{P}{2Q}ight)^{1/2}$.

Two forces of magnitude $P$ and $Q$ acting at a point have a resultant $R$. The resolved part of $R$ in the direction of $P$ is of magnitude $Q$. The angle between the forces is:

Similar Questions

$A$ person pushes a box kept on a horizontal surface with a force of $100\,N$ as shown in the figure. In unit vector notation,the force $\vec{F}$ can be expressed as?

The projection of a vector $\vec{r} = 3\hat{i} + \hat{j} + 2\hat{k}$ on the $xy$-plane has magnitude:

$A$ displacement vector has a $Y$-axis component of $10$ units. If it makes an angle of $30^{\circ}$ with the $X$-axis,find the magnitude of the vector.

Explain the resolution of vectors.

Vedclass Test Series

Exam Paper Generator

Online Exam Module