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.
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($g = 9.8 \text{ m/s}^2$)
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In the given figure the blocks $A$, $B$ and $C$ weigh 4 kg, 6 kg and 8 kg respectively. The co-efficient of sliding friction between any two surfaces is 0.5. The force $\vec{F}$ required to slide the block $C$ with constant speed is ______ N. (Use $g = 10 \text{ m/s}^2$)
