This solution explains how to find the acceleration's magnitude and direction when forces act perpendicularly on an object.
Since the forces are perpendicular, we find the magnitude of the resultant net force ($F_{net}$) using the Pythagorean theorem:
$ F_{net} = \sqrt{F_1^2 + F_2^2} $
Substituting the values:
$ F_{net} = \sqrt{(8 \text{ N})^2 + (6 \text{ N})^2} $
$ F_{net} = \sqrt{64 \text{ N}^2 + 36 \text{ N}^2} $
$ F_{net} = \sqrt{100 \text{ N}^2} $
$ F_{net} = 10 \text{ N} $
Using Newton's second law, $F_{net} = m \times a$, we can find the acceleration ($a$):
$ a = \frac{F_{net}}{m} $
Substituting the values:
$ a = \frac{10 \text{ N}}{5 \text{ kg}} $
$ a = 2 \text{ m s}^{-2} $
The acceleration is in the direction of the net force. We find the angle $\theta$ the net force makes with the $8 \text{ N}$ force.
The tangent of this angle is the ratio of the perpendicular force ($F_2$) to the force it's measured against ($F_1$):
$ \tan(\theta) = \frac{F_2}{F_1} $
$ \tan(\theta) = \frac{6 \text{ N}}{8 \text{ N}} $
$ \tan(\theta) = \frac{3}{4} $
Therefore, the angle is:
$ \theta = \tan^{-1}\left(\frac{3}{4}\right) $
The direction is $\tan^{-1}(3/4)$ with the $8 \text{ N}$ force.
The magnitude of the acceleration is $2 \text{ m s}^{-2}$ and its direction is $\tan^{-1}(3/4)$ with the $8 \text{ N}$ force.
A thin wire of length 'L' and linear mass density 'm' is bent into a circular ring (in x-y plane) with centre 'C' as shown in figure. The moment of inertia of the ring about an axis yy' (tangent in the plane) will be :
Bob B of mass $m$ at rest is hanging vertically from the ceiling via a massless string of length $10\text{ m}$, as shown in the figure. Point mass A of mass $m$ travelling horizontally with speed $10\text{ ms}^{-1}$ hits bob B elastically. The bob B rises $h$ meter after the collision. Taking the acceleration due to gravity $g = 10\text{ ms}^{-2}$ and neglecting the size of the bob, the value of $h$ is :
A thin horizontal disc is rotating about a vertical axis passing through its fixed centre O. Its angular momentum is $L_A$ and $L_B$ computed about points A and B, respectively, with $OB = 2 \times OA$. The value of $\frac{L_A}{L_B}$ is :
Two circular discs of radius each $10 \text{ cm}$ are joined at their centres by a rod of length $30 \text{ cm}$ and mass $600 \text{ gm}$ as shown in figure.
If the mass of each disc is $600 \text{ gm}$ and applied torque between two discs is $43 \times 10^5 \text{ dyne.cm}$, the angular acceleration of the discs about the given axis $AB$ is________$\text{rad/s}^2$.
